The Derivative of e^-6x

December 9, 2020October 18, 2021 ~ DerivativeIt

The derivative of e^-6x is -6e^-6x

How to calculate the derivative of e^-6x

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 e^x (the answer is e^x)
We know how to differentiate -6x (the answer is -6)

Because e^-6x is a function which is a combination of e^x and -6x, it means we can perform the differentiation of e to the -6x by making use of the chain rule.

Using the chain rule to find the derivative of e^-6x

Although the function e^-6x 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^(-6x).

Now the function is in the form of the standard exponential function e^x, except it does not have x as an exponent, instead the exponent is another function of x (-6x).

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

g(x) = -6x

From this it follows that:

e^-6x = e^g(x)

Let’s set f(x) = e^x.

Then, because g(x) = -6x, the function e^-6x can be written as a composite function of f(x) and g(x).

f(x) = e^x

f(g(x)) = e^g(x) (but g(x) = -6x)

Therefore, f(g(x)) = e^-6x

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

F(x) = f(g(x)) = e^-6x

We can now find the derivative of F(x) = e^-6x, 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 -6x.

How to find the derivative of e^-6x using the Chain Rule:

F'(x)	= f'(g(x)).g'(x)	Chain Rule Definition
	= f'(g(x))(-6)	g(x) = -6x ⇒ g'(x) = -6
	= (e^-6x)(-6)	f(g(x)) = e^-6x ⇒ f'(g(x)) = e^-6x
	= -6e^(-6x)

Using the chain rule, the derivative of e^-6x is -6e^-6x

Finally, just a note on syntax and notation: the exponential function e^-6x 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^-6x	► Derivative of e^-6x = -6e^-6x
e^(-6x)	► Derivative of e^(-6x) = -6e^-6x
e -6x	► Derivative of e -6x = -6e^-6x
e -6 x	► Derivative of e -6 x = -6e^-6x
e to the -6x	► Derivative of e to the -6x = -6e^-6x

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 e^ax = ae^ax

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

The Second Derivative of e^-6x

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

From above, we found that the first derivative of e^-6x = -6e^(-6x). So to find the second derivative of e^-6x, we just need to differentiate -6e^-6x

We can use the chain rule to calculate the derivative of -6e^-6x and get an answer of 36e^-6x.

► The second derivative of e^-6x = 36e^(-6x)

The Derivative of csc(3x)

December 9, 2020October 18, 2021 ~ DerivativeIt

The derivative of csc(3x) is -3cot(3x)csc(3x)

How to calculate the derivative of csc(3x)

In this case:

We know how to differentiate csc(x) (the answer is -csc(x)cot(x))
We know how to differentiate 3x (the answer is 3)

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

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

To perform the differentiation csc(3x), 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 3x).

Let’s call the function in the argument of csc, g(x), which means the function is in the form of csc(x), except it does not have x as the angle, instead it has another function of x (3x) as the angle

If:

g(x) = 3x

It follows that:

csc(3x) = csc(g(x))

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

f(x) = csc(x)

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

f(g(x)) = csc(3x)

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

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

We can find the derivative of csc(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 csc(3x) using the Chain Rule:

F'(x)	= f'(g(x)).g'(x)	Chain Rule Definition
	= f'(g(x))(3)	g(x) = 3x ⇒ g'(x) = 3
	= (-cot(3x)csc(3x)).(3)	f(g(x)) = csc(3x) ⇒ f'(g(x)) = -cot(3x)csc(3x)
	= -3cot(3x)csc(3x)

Using the chain rule, the derivative of csc(3x) is -3cot(3x)csc(3x)

Finally, just a note on syntax and notation: csc(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.

csc3x	► Derivative of csc3x = -3cot(3x)csc(3x)
csc 3 x	► Derivative of csc 3 x = -3cot(3x)csc(3x)
csc 3x	► Derivative of csc 3x = -3cot(3x)csc(3x)
csc (3x)	► Derivative of csc (3x) = -3cot(3x)csc(3x)
cosec(3x)	► Derivative of cosec(3x) = -3cot(3x)csc(3x)

The Second Derivative Of csc(3x)

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

From above, we found that the first derivative of csc(3x) = -3cot(3x)csc(3x). So to find the second derivative of csc(3x), we need to differentiate -3cot(3x)csc(3x).

We can use the product and chain rules, and then simplify to find the derivative of -3cot(3x)csc(3x) is 9csc³(3x) + 9cot²(3x)csc(3x)

► The second derivative of csc(3x) is 9csc³(3x) + 9cot²(3x)csc(3x)

The Derivative of tan(3x)

December 2, 2020October 18, 2021 ~ DerivativeIt

The derivative of tan(3x) is 3sec²(3x)

How to calculate the derivative of tan(3x)

In this case:

We know how to differentiate tan(x) (the answer is sec²(x))
We know how to differentiate 3x (the answer is 3)

This means the chain rule will allow us to differentiate the expression tan(3x).

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

To perform the differentiation tan(3x), 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 3x).

Let’s call the function in the argument of tan, g(x), which means the function is in the form of tan(x), except it does not have x as the angle, instead it has another function of x (3x) as the angle

If:

g(x) = 3x

It follows that:

tan(3x) = tan(g(x))

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

f(x) = tan(x)

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

f(g(x)) = tan(3x)

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

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

We can find the derivative of tan(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. But before we do that, just a quick recap on the derivative of the tan function.

The derivative of tan(x) with respect to x is sec²(x)
The derivative of tan(z) with respect to z is sec²(z)

In a similar way, the derivative of tan(3x) with respect to 3x is sec²(3x).

We will use this fact as part of the chain rule to find the derivative of tan(3x) with respect to x.

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

F'(x)	= f'(g(x)).g'(x)	Chain Rule Definition
	= f'(g(x))(3)	g(x) = 3x ⇒ g'(x) = 3
	= (sec²(3x)).(3)	f(g(x)) = tan(3x) ⇒ f'(g(x)) = sec²(3x)
	= 3sec²(3x)

Using the chain rule, the derivative of tan(3x) is 3sec²(3x)

Finally, just a note on syntax and notation: tan(3x) 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.

tan3x	► Derivative of tan3x = 3sec²(3x)
tan 3 x	► Derivative of tan 3 x = 3sec²(3x)
tan 3x	► Derivative of tan 3x = 3sec²(3x)
tan (3x)	► Derivative of tan (3x) = 3sec²(3x)

The Second Derivative Of tan(3x)

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

From above, we found that the first derivative of tan(3x) = 3sec²(3x). So to find the second derivative of tan(3x), we just need to differentiate 3sec²(3x).

We can use the chain rule to find the derivative of 3sec²(3x) (bearing in mind that the derivative of sec^2(x) is 2sec²(x)tan(x)) and it gives us a result of 18sec²(3x)tan(3x)

► The second derivative of tan(2x) is 18sec²(3x)tan(3x)

The Derivative of ln^3(x)

December 1, 2020October 18, 2021 ~ DerivativeIt

The derivative of ln^3(x) is 3ln²(x)/x

How to calculate the derivative of ln^3(x)

Note that in this post we will be looking at differentiating ln³(x) which is not the same as differentiating ln(3x). Here is our post dealing with how to differentiate ln(3x)

In this case:

We know how to differentiate ln(x) (the answer is l/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 ln^3x.

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

Although the expression ln³x contains no parenthesis, we can still view it as a composite function (a function of a function).

We can write ln³x as (ln(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 (ln(x)) as the base.

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

g(x) = ln(x)

From this it follows that:

(ln(x))³ = g(x)³

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

f(x) = x³

f(g(x)) = g(x)³ (but g(x) = ln(x))

f(g(x)) = (ln(x))³

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

F(x) = f(g(x)) = (ln(x))³

We can find the derivative of ln^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 ln^3x using the Chain Rule:

F'(x)	= f'(g(x)).g'(x)	Chain Rule Definition
	= f'(g(x))(1/x)	g(x) = ln(x) ⇒ g'(x) = 1/x
	= (3ln²(x)).(1/x))	f(g(x)) = (ln(x))³ ⇒ f'(g(x)) = 3ln²(x)
	= 3ln²(x)/x

Using the chain rule, the derivative of ln^3x is 3ln^2(x)/x

Finally, just a note on syntax and notation: ln^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.

ln³x	► Derivative of ln³x = 3ln²(x)/x
ln^3x	► Derivative of ln^3x = 3ln²(x)/x
ln 3 x	► Derivative of ln 3 x = 3ln²(x)/x
(lnx)^3	► Derivative of (lnx)^3 = 3ln²(x)/x
ln cubed x	► Derivative of ln cubed x = 3ln²(x)/x
lnx3	► Derivative of lnx3 = 3ln²(x)/x
ln^3	► Derivative of ln^3 = 3ln²(x)/x

The Second Derivative Of ln^3x

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

From above, we found that the first derivative of ln^3x = 3ln²(x)/x. So to find the second derivative of ln^3x, we just need to differentiate 3ln²(x)/x

We can use a combination of the chain rule and the quotient rule to find the derivative of 3ln²(x)/x.

We can set f(x) = ln²(x) and g(x) = x and apply the quotient rule (and the chain rule on f(x)) to find the derivative of f(x)/g(x) = 3(-ln²(x) + 2ln(x))/x²

► The second derivative of ln^3x is 3(-ln²(x) + 2ln(x))/x²

The Derivative of cot2x

December 1, 2020October 18, 2021 ~ DerivativeIt

The derivative of cot(2x) is -2csc²(2x)

How to calculate the derivative of cot(2x)

Note that in this post we will be looking at differentiating cot(2x) which is not the same as differentiating cot²(x). Here is our post dealing with how to differentiate cot^2(x).

In this case:

We know how to differentiate cot(x) (the answer is cosec²(x))
We know how to differentiate 2x (the answer is 2)

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

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

To perform the differentiation cot(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 cot, g(x), which means the function is in the form of cot(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:

cot(2x) = cot(g(x))

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

f(x) = cot(x)

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

f(g(x)) = cot(2x)

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

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

We can find the derivative of cot(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 cot function.

The derivative of cot(x) with respect to x is -csc²(x)
The derivative of cot(z) with respect to z is -csc²(z)

In a similar way, the derivative of cot(2x) with respect to 2x is -csc²(2x).

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

How to find the derivative of cot(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
	= (-csc²(2x)).(2)	f(g(x)) = cot(2x) ⇒ f'(g(x)) = -cosec²(2x)
	= -2csc²(2x)

Using the chain rule, the derivative of cot(2x) is -2csc²(2x)

Finally, just a note on syntax and notation: cot(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.

cot2x	► Derivative of cot2x = -2csc²(2x)
cot 2 x	► Derivative of cot 2 x = -2csc²(2x)
cot 2x	► Derivative of cot 2x = -2csc²(2x)
cot (2x)	► Derivative of cot (2x) = -2csc²(2x)

The Second Derivative Of cot(2x)

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

From above, we found that the first derivative of cot(2x) = -2csc²(2x). So to find the second derivative of cot(2x), we just need to differentiate -2csc²(2x).

We can use the chain rule to find the derivative of -2csc(2x) (bearing in mind that the derivative of csc^2(x) is -2csc²(x)cot(x)) and it gives us a result of 8csc²(2x)cot(2x)

► The second derivative of cot(2x) is 8csc²(2x)cot(2x)

The Derivative of ln^2(x)

December 1, 2020October 18, 2021 ~ DerivativeIt

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

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

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

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

The first method is by using the product rule for derivatives (since ln²(x) can be written as ln(x).ln(x)).

The second method is by using the chain rule for differentiation.

Finding the derivative of ln^2x using the product rule

The product rule for differentiation states that the derivative of f(x).g(x) is f’(x)g(x) + f(x).g’(x)

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

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

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

First, let F(x) = ln²(x)

Then remember that ln²(x) is equal to ln(x).ln(x)

So F(x) = ln(x)ln(x)

By setting f(x) and g(x) as ln(x) means that F(x) = f(x).g(x) and we can apply the product rule to find F'(x) (remembering that the derivative of ln(x) is 1/x)

F'(x)	= f'(x)g(x) + f(x)g'(x)	Product Rule Definition
	= f'(x)ln(x) + ln(x)g'(x)	f(x) = g(x) = ln(x)
	= (1/x)ln(x) + ln(x)(1/x)	f'(x) = g(‘x) = 1/x
	= 2ln(x)/x

Using the product rule, the derivative of ln^2x is 2ln(x)/x

Finding the derivative of ln^2x using the chain rule

In this case:

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

This means the chain rule will allow us to perform the differentiation of the expression ln^2x.

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

Although the expression ln²x contains no parenthesis, we can still view it as a composite function (a function of a function).

We can write ln²x as (ln(x))².

Now the function is in the form of x², except it does not have x as the base, instead it has another function of x (ln(x)) as the base.

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

g(x) = ln(x)

From this it follows that:

(ln(x))² = g(x)²

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

f(x) = x²

f(g(x)) = g(x)² (but g(x) = ln(x))

f(g(x)) = (ln(x))²

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

F(x) = f(g(x)) = (ln(x))²

We can find the derivative of ln^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.

How to find the derivative of ln^2x using the Chain Rule:

F'(x)	= f'(g(x)).g'(x)	Chain Rule Definition
	= f'(g(x))(1/x)	g(x) = ln(x) ⇒ g'(x) = 1/x
	= (2ln(x)).(1/x))	f(g(x)) = (ln(x))² ⇒ f'(g(x)) = 2ln(x)
	= 2ln(x)/x

Using the chain rule, the derivative of ln^2x is 2ln(x)/x

Finally, just a note on syntax and notation: ln^2x 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	► Derivative of ln²x = 2ln(x)/x
ln^2x	► Derivative of ln^2x = 2ln(x)/x
ln 2 x	► Derivative of ln 2 x = 2ln(x)/x
(lnx)^2	► Derivative of (lnx)^2 = 2ln(x)/x
ln squared x	► Derivative of ln squared x = 2ln(x)/x
lnx2	► Derivative of lnx2 = 2ln(x)/x
ln^2	► Derivative of ln^2 = 2ln(x)/x

The Second Derivative Of ln^2x

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 = 2ln(x)/x. So to find the second derivative of ln^2x, we just need to differentiate 2ln(x)/x

We can use the quotient rule to find the derivative of 2ln(x)/x.

We can set f(x) = ln(x) and g(x) = x and apply the quotient rule to find the derivative of f(x)/g(x) = 2(1-ln(x))/x²

► The second derivative of ln^2x is 2(1-ln(x))/x²

The Derivative of e^-5x

December 1, 2020October 18, 2021 ~ DerivativeIt

The derivative of e^-5x is -5e^-5x

How to calculate the derivative of e^-5x

In this case:

We know how to differentiate e^x (the answer is e^x)
We know how to differentiate -5x (the answer is -5)

Because e^-5x is a function which is a combination of e^x and -5x, it means we can perform the differentiation of e to the -5x by making use of the chain rule.

Using the chain rule to find the derivative of e^-5x

Although the function e^-5x 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^(-5x).

Now the function is in the form of the standard exponential function e^x, except it does not have x as an exponent, instead the exponent is another function of x (-5x).

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

g(x) = -5x

From this it follows that:

e^-5x = e^g(x)

Let’s set f(x) = e^x.

Then, because g(x) = -5x, the function e^-5x can be written as a composite function of f(x) and g(x).

f(x) = e^x

f(g(x)) = e^g(x) (but g(x) = -5x)

Therefore, f(g(x)) = e^-5x

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

F(x) = f(g(x)) = e^-5x

We can now find the derivative of F(x) = e^-5x, 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 -5x.

How to find the derivative of e^-5x using the Chain Rule:

F'(x)	= f'(g(x)).g'(x)	Chain Rule Definition
	= f'(g(x))(-5)	g(x) = -5x ⇒ g'(x) = -5
	= (e^-5x)(-5)	f(g(x)) = e^-5x ⇒ f'(g(x)) = e^-5x
	= -5e^(-5x)

Using the chain rule, the derivative of e^-5x is -5e^-5x

Finally, just a note on syntax and notation: the exponential function e^-5x 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^-5x	► Derivative of e^-5x = -5e^-5x
e^(-5x)	► Derivative of e^(-5x) = -5e^-5x
e -5x	► Derivative of e -5x = -5e^-5x
e -5 x	► Derivative of e -5 x = -5e^-5x
e to the -5x	► Derivative of e to the -5x = -5e^-5x

The Second Derivative of e^-5x

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

From above, we found that the first derivative of e^-5x = -5e^(-5x). So to find the second derivative of e^-5x, we just need to differentiate -5e^-5x

We can use the chain rule to calculate the derivative of -5e^-5x and get an answer of 25e^-5x.

► The second derivative of e^-5x = 25e^(-5x)

The Derivative of csc(2x)

November 25, 2020October 18, 2021 ~ DerivativeIt

The derivative of csc(2x) is -2cot(2x)csc(2x)

How to calculate the derivative of csc(2x)

Note that in this post we will be looking at differentiating csc(2x) which is not the same as differentiating csc²(x). Here is our post dealing with how to differentiate csc^2(x).

In this case:

We know how to differentiate csc(x) (the answer is -csc(x)cot(x))
We know how to differentiate 2x (the answer is 2)

This means the chain rule will allow us to perform the differentiation of the expression csc(2x).

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

To perform the differentiation csc(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 csc, g(x), which means the function is in the form of csc(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:

csc(2x) = csc(g(x))

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

f(x) = csc(x)

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

f(g(x)) = csc(2x)

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

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

We can find the derivative of csc(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.

How to find the derivative of csc(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
	= (-cot(2x)csc(2x)).(2)	f(g(x)) = csc(2x) ⇒ f'(g(x)) = -cot(2x)csc(2x)
	= -2cot(2x)csc(2x)

Using the chain rule, the derivative of csc(2x) is -2cot(2x)csc(2x)

Finally, just a note on syntax and notation: csc(2x) 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.

csc2x	► Derivative of csc2x = -2cot(2x)csc(2x)
csc 2 x	► Derivative of csc 2 x = -2cot(2x)csc(2x)
csc 2x	► Derivative of csc 2x = -2cot(2x)csc(2x)
csc (2x)	► Derivative of csc (2x) = -2cot(2x)csc(2x)
cosec(2x)	► Derivative of cosec(2x) = -2cot(2x)csc(2x)

The Second Derivative Of csc(2x)

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

From above, we found that the first derivative of csc(2x) = -2cot(2x)csc(2x). So to find the second derivative of csc(2x), we need to differentiate -2cot(2x)csc(2x).

We can use the product and chain rules, and then simplify to find the derivative of -2cot(2x)csc(2x) is 4csc³(2x) + 4cot²(2x)csc(2x)

► The second derivative of csc(2x) is 4csc³(2x) + 4cot²(2x)csc(2x)

The Derivative of sec2x

November 13, 2020October 18, 2021 ~ 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).

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)

November 13, 2020October 18, 2021 ~ 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)).1	f(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