Author: DerivativeIt

The Derivative of sec2x

The derivative of sec(2x) is 2sec(2x)tan(2x)
~ DerivativeIt

The derivative of sec(2x) is 2sec(2x)tan(2x)

How to calculate the derivative of sec(2x)

Note that in this post we will be looking at differentiating sec(2x) which is not the same as differentiating sec2(x). Here is our post dealing with how to differentiate sec^2(x).

The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate sec(x) (the answer is sec(x)tan(x))
  • We know how to differentiate 2x (the answer is 2)

This means the chain rule will allow us to differentiate the expression sec(2x).

Using the chain rule to find the derivative of sec(2x)

To perform the differentiation sec(2x), the chain rule says we must differentiate the expression as if it were just in terms of x as long as we then multiply that result by the derivative of what the expression is actually in terms of (in this case the derivative of 2x).

Let’s call the function in the argument of sec, g(x), which means the function is in the form of sec(x), except it does not have x as the angle, instead it has another function of x (2x) as the angle

If:

g(x) = 2x

It follows that:

sec(2x) = sec(g(x))

So if the function f(x) = sec(x) and the function g(x) = 2x, then the function sec(2x) can be written as a composite function.

f(x) = sec(x)

f(g(x)) = sec(g(x)) (but g(x) = 2x))

f(g(x)) = sec(2x)

Let’s define this composite function as F(x):

F(x) = f(g(x)) = sec(2x)

We can find the derivative of sec(2x) (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)


If F(x) = f(g(x))


Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule. But before we do that, just a quick recap on the derivative of the sec function.

The derivative of sec(x) with respect to x is sec(x)tan(x)
The derivative of sec(z) with respect to z is sec(z)tan(z)

In a similar way, the derivative of sec(2x) with respect to 2x is sec(2x)tan(2x).

We will use this fact as part of the chain rule to find the derivative of sec(2x) with respect to x.

How to find the derivative of sec(2x) using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x))(2)g(x) = 2x ⇒ g'(x) = 2
= (sec(2x)tan(2x)).(2)f(g(x)) = sec(2x) ⇒ f'(g(x)) = sec(2x)tan(2x)
= 2sec(2x)tan(2x)

Using the chain rule, the derivative of sec(2x) is 2sec(2x)tan(2x)

Finally, just a note on syntax and notation: sec(2x) is sometimes written in the forms below (with the derivative as per the calculation above). Just be aware that not all of the forms below are mathematically correct.

sec2x► Derivative of sec2x = 2sec(2x)tan(2x)
sec 2 x► Derivative of sec 2 x = 2sec(2x)tan(2x)
sec 2x► Derivative of sec 2x = 2sec(2x)tan(2x)
sec (2x)► Derivative of sec (2x) = 2sec(2x)tan(2x)

The Second Derivative Of sec(2x)

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of sec(2x) = 2sec(2x)tan(2x). So to find the second derivative of sec(2x), we just need to differentiate 2sec(2x)tan(2x)

We can use the chain rule to find the derivative of 2sec(2x)tan(2x) and it gives us a result of 8sec3(2x) – 4sec(2x)

► The second derivative of sec(2x) is 8sec3(2x) – 4sec(2x)

The Derivative of ln(x+1)

The derivative of ln(x+1) is 1/(x+1)
~ DerivativeIt

The derivative of ln(x+1) is 1/(x+1)

How to calculate the derivative of ln(x+1)

The chain rule is useful for finding the derivative of an expression which could have been differentiated had it been in terms of x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate x+1 (the answer is 1)
  • We know how to differentiate ln(x) (the answer is 1/x)

This means the chain rule will allow us to perform the differentiation of the function ln(x+1).

To perform the differentiation, the chain rule says we must differentiate the expression as if it were just in terms of x as long as we then multiply that result by the derivative of what the expression was actually in terms of (in this case the derivative of x+1).

Using the chain rule to find the derivative of ln(x+1)

ln(x+1) is in the form of the standard natural log function ln(x), except it does not have x as an argument, instead it has another function of x (x+1).

Let’s call the function in the argument g(x), which means:

g(x) = x+1

From this it follows that:

ln(x+1) = ln(g(x))

So if the function f(x) = ln(x) and the function g(x) = x+1, then the function ln(x+1) can be written as a composite function.

f(x) = ln(x)

f(g(x)) = ln(g(x)) (but g(x) = x+1)

f(g(x)) = ln(x+1)

Let’s define this composite function as F(x):

F(x) = f(g(x)) = ln(x+1)

We can find the derivative of ln(x+1) (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule. But before we do that, just a recap on the derivative of the natural logarithm.

The derivative of ln(x) with respect to x is (1/x)
The derivative of ln(s) with respect to s is (1/s)

In a similar way, the derivative of ln(x+1) with respect to x+1 is 1/(x+1).
We will use this fact as part of the chain rule to find the derivative of ln(x+1) with respect to x.

How to find the derivative of ln(x+1) using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x)).(1)g(x) = x+1 ⇒ g'(x) = 1
= (1/(x+1)).1f(g(x)) = ln(x+1) ⇒ f'(g(x)) = 1/(x+1)
(The derivative of ln(x+1) with respect to x+1 is 1/(x+1)
= 1/(x+1)

Using the chain rule, we find that the derivative of ln(x+1) is 1/(x+1)

Finally, just a note on syntax and notation: ln(x+1) is sometimes written in the forms below (with the derivative as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

lnx+1► Derivative of lnx+1 =1/(x+1)
ln x+1► Derivative of ln x+1 = 1/(x+1)
ln x + 1► Derivative of ln x +1 = 1/(x+1)

The Second Derivative of ln(x+1)

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of ln(x+1) = 1/(x+1). So to find the second derivative of ln(x+1), we just need to differentiate 1/(x+1).

We can use the quotient rule to find the derivative of 1/(x+1), and we get an answer of -1/(x+1)2

► The second derivative of ln(x+1) = -1/(x+1)2

The Derivative of lnx^2

The derivative of ln(x^2) is 2/x
~ DerivativeIt

The derivative of ln(x^2) is 2/x

How to calculate the derivative of lnx^2

Note that in this post we will be looking at differentiating ln(x2) which is not the same as differentiating ln2(x) or ln(2x). Here are our posts dealing with how to differentiate ln2(x) and how to differentiate ln(2x)

There are two methods that can be used for calculating the derivative of ln(x2).

The first method is by using the chain rule for derivatives.

The second method is by using the properties of logs to write ln(x2) into a form which differentiable without needing to use the chain rule.

Finding the derivative of ln(x2) using the chain rule

The chain rule is useful for finding the derivative of an expression which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate x2(the answer is 2x)
  • We know how to differentiate ln(x) (the answer is 1/x)

This means the chain rule will allow us to perform the differentiation of the function ln(x2).

To perform the differentiation, the chain rule says we must differentiate the expression as if it were just in terms of x as long as we then multiply that result by the derivative of what the expression was actually in terms of (in this case the derivative of x2).

Using the chain rule to find the derivative of ln(x^2)

ln(x2) is in the form of the standard natural log function ln(x), except it does not have x as an argument, instead it has another function of x (x2).

Let’s call the function in the argument g(x), which means:

g(x) = x2

From this it follows that:

ln(x2) = ln(g(x))

So if the function f(x) = ln(x) and the function g(x) = x2, then the function ln(x2) can be written as a composite function.

f(x) = ln(x)

f(g(x)) = ln(g(x)) (but g(x) = x2)

f(g(x)) = ln(x2)

Let’s define this composite function as F(x):

F(x) = f(g(x)) = ln(x2)

We can find the derivative of ln(x2) (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule. But before we do that, just a recap on the derivative of the natural logarithm.

The derivative of ln(x) with respect to x is (1/x)
The derivative of ln(s) with respect to s is (1/s)

In a similar way, the derivative of ln(x2) with respect to x2 is (1/x2).
We will use this fact as part of the chain rule to find the derivative of ln(x2) with respect to x.

How to find the derivative of ln(x2) using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x)).(2x)g(x) = x2⇒ g'(x) = 2x
= (1/x2).2xf(g(x)) = ln(x2) ⇒ f'(g(x)) = 1/x2
(The derivative of ln(x2) with respect to x2is (1/x2))
= 2/x

Using the chain rule, we find that the derivative of ln(x2) is 2/x

Finally, just a note on syntax and notation: ln(x^2) is sometimes written in the forms below (with the derivative as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

ln(x2)► Derivative of ln(x2) = 2/x
lnx2► Derivative of lnx2 = 2/x
lnx^2► Derivative of lnx^2 = 2/x
ln x 2► Derivative of ln x 2 = 2/x
ln x squared► Derivative of ln x squared = 2/x

Finding the derivative of ln(x2) using log properties

Since ln is the natural logarithm, the usual properties of logs apply.

The power property of logs states that ln(xy) = y.ln(x). In other words taking the log of x to a power is the same as multiplying the log of x by that power.

We can therefore use the power rule of logs to rewrite ln(x2) as:

f(x) = ln(x2) = 2.ln(x)

How to find the derivative of ln(x2) using the power rule of logs

f(x)= 2ln(x)
f'(x)= 2.(1/x)The derivative of ln(x) is 1/x
= 2/x

The Second Derivative of ln(x2)

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of ln(x2) = 2/x. So to find the second derivative of ln(x2), we just need to differentiate 2/x

If we differentiate 2/x we get an answer of (-2/x2).

► The second derivative of ln(x2) = -2/x2

The Derivative of e^x^3

The derivative of e^x^3 is (3x^2)(e^x^3)
~ DerivativeIt

The derivative of e^x^3 is (3x^2)(e^x^3)

How to calculate the derivative of e^x^3

The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate ex (the answer is ex)
  • We know how to differentiate x3 (the answer is 3x)

Because e^x^3 is a function which is a combination of ex and x3, it means we can perform the differentiation of e to the x3 by making use of the chain rule.

Using the chain rule to find the derivative of e^x^3

Although the function ex3 contains no parenthesis, we can still view it as a composite function (a function of a function).

If we add parenthesis around the exponent, we get e(x3).

Now the function is in the form of the standard exponential function ex, except it does not have x as an exponent, instead the exponent is another function of x (x3).

Let’s call the function in the exponent g(x), which means:

g(x) = x3

From this it follows that:

ex3 = eg(x)

Let’s set f(x) = ex.

Then, because g(x) = x3, the function ex3 can be written as a composite function of f(x) and g(x).

f(x) = ex

f(g(x)) = eg(x) (but g(x) = x3)

Therefore, f(g(x)) = ex3

Let’s define this composite function as F(x):

F(x) = f(g(x)) = ex3

We can now find the derivative of F(x) = e^x^3, F'(x), by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule to find the derivative of e to the x squared.

How to find the derivative of e^x^3 using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x))(3x2)g(x) = x3 ⇒ g'(x) = 3x2
= (e^x^3).(3x2)f(g(x)) = e^x^3 ⇒ f'(g(x)) = e^x^3
= (3x2)ex3

Using the chain rule, the derivative of e^x^3 is (3x2)ex3

Finally, just a note on syntax and notation: the exponential function e^x^3 is sometimes written in the forms shown below (the derivative of each is as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

ex3► Derivative of ex3 =(3x2)ex3
e^(x^3)► Derivative of e^(x^3) = (3x2)ex3
e x 3► Derivative of e x 3 = (3x2)ex3
e to the x cubed► Derivative of e to the x cubed = (3x2)ex3

The Second Derivative of e^x^3

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of e^x^3 = (3x2)ex3. So to find the second derivative of e^x^3, we just need to differentiate (3x2)ex3.

We can use the chain rule in combination with the product rule for differentiation to calculate the derivative of (3x2)ex3. The answer is (6x)ex3 + (9x4)ex3

► The second derivative of e^x^3 = (6x)ex3 + (9x4)ex3

The Derivative of ln(2x^2)

The derivative of ln(2x^2) is 2/x
~ DerivativeIt

The derivative of ln(2x^2) is 2/x

How to calculate the derivative of ln(2x^2)

There are two methods that can be used for calculating the derivative of ln(2x2).

The first method is by using the chain rule for derivatives.

The second method is by using the properties of logs to write ln(2x2) into a form which differentiable without needing to use the chain rule.

Finding the derivative of ln(2x2) using the chain rule

The chain rule is useful for finding the derivative of an expression which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate 2x2(the answer is 4x)
  • We know how to differentiate ln(x) (the answer is 1/x)

This means the chain rule will allow us to perform the differentiation of the function ln(2x2).

To perform the differentiation, the chain rule says we must differentiate the expression as if it were just in terms of x as long as we then multiply that result by the derivative of what the expression was actually in terms of (in this case the derivative of 2x2).

Using the chain rule to find the derivative of ln(2x^2)

ln(2x2) is in the form of the standard natural log function ln(x), except it does not have x as an argument, instead it has another function of x (2x2).

Let’s call the function in the argument g(x), which means:

g(x) = 2x2

From this it follows that:

ln(2x2) = ln(g(x))

So if the function f(x) = ln(x) and the function g(x) = 2x2, then the function ln(2x2) can be written as a composite function.

f(x) = ln(x)

f(g(x)) = ln(g(x)) (but g(x) = 2x2)

f(g(x)) = ln(2x2)

Let’s define this composite function as F(x):

F(x) = f(g(x)) = ln(2x2)

We can find the derivative of ln(2x2) (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule. But before we do that, just a recap on the derivative of the natural logarithm.

The derivative of ln(x) with respect to x is (1/x)
The derivative of ln(s) with respect to s is (1/s)

In a similar way, the derivative of ln(2x2) with respect to 2x2 is (1/2x2).
We will use this fact as part of the chain rule to find the derivative of ln(2x2) with respect to x.

How to find the derivative of ln(2x2) using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x)).(4x)g(x) = 2x2⇒ g'(x) = 4x
= (1/2x2).4xf(g(x)) = ln(2x2) ⇒ f'(g(x)) = 1/2x2
(The derivative of ln(2x2) with respect to 2x2is (1/2x2))
= 2/x

Using the chain rule, we find that the derivative of ln(2x2) is 2/x

Finally, just a note on syntax and notation: ln(2x^2) is sometimes written in the forms below (with the derivative as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

ln2x2► Derivative of ln2x2 = 2/x
ln2x^2► Derivative of ln2x^2 = 2/x
ln 2x 2► Derivative of ln 2x 2 = 2/x
ln 2 x squared► Derivative of ln 2 x squared = 2/x

Finding the derivative of ln(2x2) using log properties

Since ln is the natural logarithm, the usual properties of logs apply.

The product property of logs states that ln(xy) = ln(x) + ln(y). In other words taking the log of a product is equal to the summing the logs of each term of the product.

Since 2x2 is the product of 2 and x2, we can use the product properties of logs to rewrite ln(2x2):

f(x) = ln(2x2) = ln(2) + ln(x2)

The power property of logs states that ln(xy) = y.ln(x). In other words taking the log of x to a power is the same as multiplying the log of x by that power.

We can therefore combine the product and power rules of logs to rewrite ln(2x2) as:

f(x) = ln(2x2) = ln(2) + ln(x2) = ln(2) + 2.ln(x)

How to find the derivative of ln(2x2) using the product property of logs

f(x)= ln(2) + 2ln(x)
f'(x)= 0 + ln(x)ln2 is a constant, the derivative of a constant is 0
= 0 + 2/xThe derivative of ln(x) is 1/x
= 2/x

The Second Derivative of ln(2x2)

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of ln(2x2) = 2/x. So to find the second derivative of ln(2x2), we just need to differentiate 2/x

If we differentiate 2/x we get an answer of (-2/x2).

► The second derivative of ln(2x2) = -2/x2

The Derivative of cos^3x

The derivative of cos^3x is equal to -3cos^2(x)sin(x)
~ DerivativeIt

The derivative of cos^3x is -3cos2(x)sin(x)

How to calculate the derivative of cos^3x

Note that in this post we will be looking at differentiating cos3(x) which is not the same as differentiating cos(3x). Here is our post dealing with how to differentiate cos(3x).

The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate cos(x) (the answer is -sin(x))
  • We know how to differentiate x3 (the answer is 3x2)

This means the chain rule will allow us to perform the differentiation of the expression cos3(x).

Using the chain rule to find the derivative of cos^3x

Although the expression cos3x contains no parenthesis, we can still view it as a composite function (a function of a function).

We can write cos3x as (cos(x))3.

Now the function is in the form of x3, except it does not have x as the base, instead it has another function of x (cos(x)) as the base.

Let’s call the function of the base g(x), which means:

g(x) = cos(x)

From this it follows that:

cos(x)3 = g(x)3

So if the function f(x) = x3 and the function g(x) = cos(x), then the function (cos(x))3 can be written as a composite function.

f(x) = x3

f(g(x)) = g(x)3 (but g(x) = cos(x))

f(g(x)) = (cos(x))3

Let’s define this composite function as F(x):

F(x) = f(g(x)) = (cos(x))3

We can find the derivative of cos^3x (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)


If F(x) = f(g(x))


Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule.

How to find the derivative of cos^3x using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x))(-sin(x))g(x) = cos(x) ⇒ g'(x) = sin(x)
= (3cos2(x)).(-sin(x))f(g(x)) = (cos(x))3 ⇒ f'(g(x)) = 3cos2(x)
= -3cos2(x)sin(x)

Using the chain rule, the derivative of cos^3x is -3cos2(x)sin(x)

Finally, just a note on syntax and notation: cos^3x is sometimes written in the forms below (with the derivative as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

cos3x► Derivative of cos3x = -3cos2(x)sin(x)
cos^3(x)► Derivative of cos^3(x) = -3cos2(x)sin(x)
cos 3 x► Derivative of cos 3 x = -3cos2(x)sin(x)
(cosx)^3► Derivative of (cosx)^3 = -3cos2(x)sin(x)
cos cubed x► Derivative of cos cubed x = -3cos2(x)sin(x)
cos x cubed► Derivative of cos x cubed = -3cos2(x)sin(x)
cosx3► Derivative of cosx3 = -3cos2(x)sin(x)
cos^3► Derivative of cos^3 = -3cos2(x)sin(x)

The Second Derivative Of cos^3x

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of cos^3x = -3cos2(x)sin(x). So to find the second derivative of cos^3x, we just need to differentiate -3cos2(x)sin(x).

We can use the product rule and trig identities to find the derivative of -3cos2(x)sin(x) and we get an answer of -3cos3(x) + 3sin(x)sin(2x)

► The second derivative of cos^3x is -3cos3(x) + 3sin(x)sin(2x)

The Derivative of e^-4x

The derivative of e^-4x is equal to -4e^-4x
~ DerivativeIt

The derivative of e^-4x is -4e^-4x

How to calculate the derivative of e^-4x

The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate ex (the answer is ex)
  • We know how to differentiate -4x (the answer is -4)

Because e^-4x is a function which is a combination of ex and -4x, it means we can perform the differentiation of e to the -4x by making use of the chain rule.

Using the chain rule to find the derivative of e^-4x

Although the function e-4x contains no parenthesis, we can still view it as a composite function (a function of a function).

If we add parenthesis around the exponent, we get e(-4x).

Now the function is in the form of the standard exponential function ex, except it does not have x as an exponent, instead the exponent is another function of x (-4x).

Let’s call the function in the exponent g(x), which means:

g(x) = -4x

From this it follows that:

e-4x = eg(x)

Let’s set f(x) = ex.

Then, because g(x) = -4x, the function e-4x can be written as a composite function of f(x) and g(x).

f(x) = ex

f(g(x)) = eg(x) (but g(x) = -4x)

Therefore, f(g(x)) = e-4x

Let’s define this composite function as F(x):

F(x) = f(g(x)) = e-4x

We can now find the derivative of F(x) = e^-4x, F'(x), by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule to find the derivative of e to the -4x.

How to find the derivative of e^-4x using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x))(-4)g(x) = -4x ⇒ g'(x) = -4
= (e^-4x)(-4)f(g(x)) = e^-4x ⇒ f'(g(x)) = e^-4x
= -4e^(-4x)

Using the chain rule, the derivative of e^-4x is -4e^-4x

Finally, just a note on syntax and notation: the exponential function e^-4x is sometimes written in the forms shown below (the derivative of each is as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

e-4x► Derivative of e-4x = -4e-4x
e^(-4x)► Derivative of e^(-4x) = -4e-4x
e -4x► Derivative of e -4x = -4e-4x
e -4 x► Derivative of e -4 x = -4e-4x
e to the -4x► Derivative of e to the -4x = -4e-4x

Top Tip

It’s possible to generalize the derivative of expressions in the form e^ax (where a is a constant value):

The derivative of eax = aeax

(Add the constant a to the front of the expression and keep the exponential part the same)

The Second Derivative of e^-4x

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of e^-4x = -4e^(-4x). So to find the second derivative of e^-4x, we just need to differentiate -4e-4x

We can use the chain rule to calculate the derivative of -4e-4x and get an answer of 16e-4x.

► The second derivative of e^-4x = 16e^(-4x)

The Derivative of sin^3(x)

The derivative of sin^3x is 3sin^2(x)cos(x)
~ DerivativeIt

The derivative of sin^3(x) is 3sin^2(x)cos(x)

How to calculate the derivative of sin^3x

Note that in this post we will be looking at differentiating sin3(x) which is not the same as differentiating sin(3x). Here is our post dealing with how to differentiate sin(3x).

The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate sin(x) (the answer is cos(x))
  • We know how to differentiate x3 (the answer is 3x2)

This means the chain rule will allow us to perform the differentiation of the expression sin^3(x).

Using the chain rule to find the derivative of sin^3(x)

Although the expression sin3x contains no parenthesis, we can still view it as a composite function (a function of a function).

We can write sin3x as (sin(x))3.

Now the function is in the form of x3, except it does not have x as the base, instead it has another function of x (sin(x)) as the base.

Let’s call the function of the base g(x), which means:

g(x) = sin(x)

From this it follows that:

sin(x)3 = g(x)3

So if the function f(x) = x3 and the function g(x) = sin(x), then the function (sin(x))3 can be written as a composite function.

f(x) = x3

f(g(x)) = g(x)3 (but g(x) = sin(x))

f(g(x)) = (sin(x))3

Let’s define this composite function as F(x):

F(x) = f(g(x)) = (sin(x))3

We can find the derivative of sin^3x (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)


If F(x) = f(g(x))


Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule.

How to find the derivative of sin^3(x) using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x))(cos(x))g(x) = sin(x) ⇒ g'(x) = cos(x)
= (3sin2(x)).(cos(x))f(g(x)) = (sin(x))3 ⇒ f'(g(x)) = 3sin2(x)
= 3sin2(x)cos(x)

Using the chain rule, the derivative of sin^3(x) is 3sin2(x)cos(x)

Finally, just a note on syntax and notation: sin^3(x) is sometimes written in the forms below (with the derivative as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

sin3x► Derivative of sin3x = 3sin2(x)cos(x)
sin^3x► Derivative of sin^3x = 3sin2(x)cos(x)
sin 3 x► Derivative of sin 3 x = 3sin2(x)cos(x)
(sinx)^3► Derivative of (sinx)^3 = 3sin2(x)cos(x)
sin cubed x► Derivative of sin cubed x = 3sin2(x)cos(x)
sin x cubed► Derivative of sin x cubed = 3sin2(x)cos(x)
sinx3► Derivative of sinx3 = 3sin2(x)cos(x)
sin^3► Derivative of sin^3 = 3sin2(x)cos(x)

The Second Derivative Of sin^3(x)

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of sin^3x = 3sin2(x)cos(x). So to find the second derivative of sin^2x, we just need to differentiate 3sin2(x)cos(x).

We can use the product rule and trig identities to find the derivative of 3sin2(x)cos(x). We get an answer of -3sin3x + 3cos(x)sin(2x)

► The second derivative of sin^3(x) is -3sin3x + 3cos(x)sin(2x)

The Derivative of ln(2x+1)

The derivative of ln(2x+1) is 2/(2x+1)
~ DerivativeIt

The derivative of ln(2x+1) is 2/(2x+1)

How to calculate the derivative of ln(2x+1)

The chain rule is useful for finding the derivative of an expression which could have been differentiated had it been in terms of x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate 2x+1 (the answer is 2)
  • We know how to differentiate ln(x) (the answer is 1/x)

This means the chain rule will allow us to perform the differentiation of the function ln(2x+1).

To perform the differentiation, the chain rule says we must differentiate the expression as if it were just in terms of x as long as we then multiply that result by the derivative of what the expression was actually in terms of (in this case the derivative of 2x+1).

Using the chain rule to find the derivative of ln(2x+1)

ln(2x+1) is in the form of the standard natural log function ln(x), except it does not have x as an argument, instead it has another function of x (2x+1).

Let’s call the function in the argument g(x), which means:

g(x) = 2x+1

From this it follows that:

ln(2x+1) = ln(g(x))

So if the function f(x) = ln(x) and the function g(x) = 2x+1, then the function ln(2x+1) can be written as a composite function.

f(x) = ln(x)

f(g(x)) = ln(g(x)) (but g(x) = 2x+1)

f(g(x)) = ln(2x+1)

Let’s define this composite function as F(x):

F(x) = f(g(x)) = ln(2x+1)

We can find the derivative of ln(2x+1) (F'(x)) by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule. But before we do that, just a recap on the derivative of the natural logarithm.

The derivative of ln(x) with respect to x is (1/x)
The derivative of ln(s) with respect to s is (1/s)

In a similar way, the derivative of ln(2x+1) with respect to 2x+1 is 1/(2x+1).
We will use this fact as part of the chain rule to find the derivative of ln(2x+1) with respect to x.

How to find the derivative of ln(2x+1) using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x)).(2)g(x) = 2x+1 ⇒ g'(x) = 2
= (1/(2x+1)).2f(g(x)) = ln(2x+1) ⇒ f'(g(x)) = 1/(2x+1)
(The derivative of ln(2x+1) with respect to 2x+1 is 1/(2x+1)
= 2/(2x+1)

Using the chain rule, we find that the derivative of ln(2x+1) is 2/(2x+1)

Finally, just a note on syntax and notation: ln(2x+1) is sometimes written in the forms below (with the derivative as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

ln2x+1► Derivative of ln2x+1 =2/(2x+1)
ln 2x+1► Derivative of ln 2x+1 = 2/(2x+1)
ln 2 x + 1► Derivative of ln 2 x +1 = 2/(2x+1)

The Second Derivative of ln(2x+1)

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of ln(2x+1) = 2/(2x+1). So to find the second derivative of ln(2x+1), we just need to differentiate 2/(2x+1).

We can use the quotient rule to find the derivative of 2/(2x+1), and we get an answer of -4/(2x+1)2

► The second derivative of ln(2x+1) = -4/(2x+1)2

The Derivative of e^-x

The derivative of e^-x is equal to -e^-x
~ DerivativeIt

The derivative of e^-x is -e^-x

How to calculate the derivative of e^-x

The chain rule is useful for finding the derivative of a function which could have been differentiated had it been in x, but it is in the form of another expression which could also be differentiated if it stood on its own.

In this case:

  • We know how to differentiate ex (the answer is ex)
  • We know how to differentiate -x (the answer is -1)

Because e^-x is a function which is a combination of ex and -x, it means we can perform the differentiation of e to the -x by making use of the chain rule.

Using the chain rule to find the derivative of e^-x

Although the function e-x contains no parenthesis, we can still view it as a composite function (a function of a function).

If we add parenthesis around the exponent, we get e(-x).

Now the function is in the form of the standard exponential function ex, except it does not have x as an exponent, instead the exponent is another function of x (-x).

Let’s call the function in the exponent g(x), which means:

g(x) = -x

From this it follows that:

e-x = eg(x)

Let’s set f(x) = ex.

Then, because g(x) = -x, the function e-x can be written as a composite function of f(x) and g(x).

f(x) = ex

f(g(x)) = eg(x) (but g(x) = -x)

Therefore, f(g(x)) = e-x

Let’s define this composite function as F(x):

F(x) = f(g(x)) = e-x

We can now find the derivative of F(x) = e^-x, F'(x), by making use of the chain rule.

The Chain Rule:
For two differentiable functions f(x) and g(x)

If F(x) = f(g(x))

Then the derivative of F(x) is F'(x) = f’(g(x)).g’(x)

Now we can just plug f(x) and g(x) into the chain rule to find the derivative of e to the negative x.

How to find the derivative of e^-x using the Chain Rule:

F'(x)= f'(g(x)).g'(x)Chain Rule Definition
= f'(g(x))(-1)g(x) = -x ⇒ g'(x) = –1
= (e^-x).(-1)f(g(x)) = e^-x ⇒ f'(g(x)) = e^-x
= -e^(-x)

Using the chain rule, the derivative of e^-x is -e^-x

Finally, just a note on syntax and notation: the exponential function e^-x is sometimes written in the forms shown below (the derivative of each is as per the calculations above). Just be aware that not all of the forms below are mathematically correct.

e-x► Derivative of e-x = -e-x
e^(-x)► Derivative of e^(-x) = -e-x
e -x► Derivative of e -x = -e-x
e – x► Derivative of e – x = -e-x
e to the negative x► Derivative of e to the negative x = -e-x
e to the -x► Derivative of e to the -x = -e-x

Top Tip

It’s possible to generalize the derivative of expressions in the form e^ax (where a is a constant value):

The derivative of eax = aeax

(Add the constant a to the front of the expression and keep the exponential part the same)

The Second Derivative of e^-x

To calculate the second derivative of a function, you just differentiate the first derivative.

From above, we found that the first derivative of e^-x = -e^(-x). So to find the second derivative of e^-x, we just need to differentiate -e-x

We can use the chain rule to calculate the derivative of -e-x and get an answer of e-x (i.e. the second derivative of e-x is just itself).

► The second derivative of e^-x = e^(-x)