In calculus, the derivative of cos(x) is equal to -sin(x)

(d/dx)cos(x) = -sin(x)

Or

cos'(x) = -sin(x)

So what does this mean and how do we get to that answer?

Well, since differentiation involves finding the rate of change (or gradient) of a function, finding the derivative of cos means that we are trying to determine the rate of change of the cos function with respect to x (the angle of the cos function).

This evaluation of the rate of change of the cos function with respect to the variable x is found to be negative sine of x (-sin(x)).

Next we will look at how we calculate the derivative of cos(x).

The derivative of cos(x) from first principles

To find the derivative of cos(x) (or in fact any function) it is often a good idea to start with first principles and the definition of the derivative of a function.

Recall that derivative of a function f(x) (or f'(x)) can be expressed as

In this case f(x), the function we want to differentiate is cos(x). So if we put f(x) = cos(x) into the definition of the derivative function we get

Next, we need to use the following trigonomic identity relating to cos and the sum of two angles. You may recall (if you don’t then just accept it as fact for now)

We can then apply this trig identity to the cos(x+h) term from above to get:

Then we can do some factorisation of the numerator by rearranging it a little and taking out a common cos(x) term:

Finally, we split the numerator up so that we can apply two formulas related to finding the limits of trigonometric functions.

The following two formulas relating to limits of sin and cos now come in handy. We can plug these two in and then solve:

And that proves that the derivative of cos(x) is -sin(x)

The product rule for calculus is useful for finding the derivative of a function which is expressed as the product of two differentiable functions.

Say for example we had two functions

f(x) = x and g(x) = x^{2}

Now say we wanted to find the derivative of

One approach to finding the derivative would be to simplify the function by multiplying it, and then differentiate the answer. Like this:

So the derivative of the product f(x).g(x) is 3x^{2} (since the derivative of x^{3} is 3x^{2}.

Next, let’s calculate the derivative of f'(x).g'(x). In other words let’s differentiate each function first, and then multiply the derivatives together.

From the above we can see that (f(x)g(x))’ = 3x^{2}, and that is not equal to f'(x)g'(x) = 2x

The derivative of a product is not equal to the product of the derivatives

To differentiate a product, you cannot just take the derivative of the first function and multiply it by the derivative of the second function. Instead, we need to use the product rule to find the derivative of a product (in a similar way that the quotient rule needs to be used to find the derivative of a quotient).

The Product Rule Formula For Differentiation

If two functions f(x) and g(x) are differentiable (i.e. the derivatives of f(x) and g(x) exist), then their product f(x).g(x) is differentiable, and the derivative can be found as follows:

The formula for the product rule in words

The Product Rule states that the derivative of a product is equal to the derivative of the first function times the second function plus the first function times the derivative of the second function.

When To Use The Product Rule

In calculus, the product rule can be used when the function you want to differentiate consists of a product, and both parts of the product are differentiable functions of their own.

For example the function f(x) = x.x^{2} can be differentiated using the product rule for derivatives because:

The first function (x) is differentiable – its derivative is 1

The second function (x^{2}) is differentiable – it’s derivative is 2x

(Of course you could also first simplify the function to x^{3} and then differentiate it and get the same result)

Examples Using The Product Rule

The best way to understand how the product rule works is by looking at some examples.

Using the product rule

For the first example, let’s use the product rule to find the derivative of

The above function is actually a product of two other functions, namely x and cos(x)

We can call these two functions f(x) and g(x): f(x) = x g(x) = cos(x)

And the derivatives of these two functions are: f'(x) = 1 (since the derivative of x is 1) g'(x) = -sin(x) (since the derivative of cos(x) is -sin(x)

We can now apply the product rule to find the derivative of the function xcos(x) which is a product of f(x) and g(x).

The formula for the product rule is:

Next we plug f(x), f'(x), g(x) and g'(x) into the above formula:

And then we can simplify the above to get a final answer of:

Proof Of The Product Rule

Proving the product rule using first principles

Let F(x) = f(x)g(x)

The definition of the derivative of F(x) is

If we insert F(x) = f(x)g(x) into the definition we get:

This does not help us much in terms of simplification, so we need to pull a little trick to take this further.

What we do is we add 0 to the numerator, which of course does not change the value at all. The trick is to add 0 in the form of f(x+h)g(x) – f(x+h)g(x) instead of adding a straight 0. Note we also swap the order around (-f(x+h)g(x) + f(x+h)g(x)) to make things easier.

Next we use one of the properties of limits which states that the limit of a sum is equal to the sum of the limits. So we can split the above limit into two parts. Like this:

Now we can perform factorisation by taking out the term f(x+h) in the first numerator and g(x) in the second numerator.

We now again apply a property of limits which says that the limit of a product of two functions is equal to the product of their limits. So each of the two terms above can be written as the products of two limits.

Now if we examine each of these limits individually:

Plugging these individual solutions to each of the limits from above gives us the final version of the product rule.

The Product Rule in Terms of u and v

With regards to notation, the product rule is sometimes easier to express if you use u and v to represent the two functions which make up the product. This can also be more compact than using the function notation f(x) and g(x).

Using the variable u for the first function of the product, and v for the second function of the product, the product rule for finding the derivative of the function uv can be expressed as:

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 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 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 9csc^{3}(3x) + 9cot^{2}(3x)csc(3x)

► The second derivative of csc(3x) is 9csc^{3}(3x) + 9cot^{2}(3x)csc(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 tan(x) (the answer is sec^{2}(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 sec^{2}(x) The derivative of tan(z) with respect to z is sec^{2}(z)

In a similar way, the derivative of tan(3x) with respect to 3x is sec^{2}(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

= (sec^{2}(3x)).(3)

f(g(x)) = tan(3x) ⇒f'(g(x)) = sec^{2}(3x)

= 3sec^{2}(3x)

Using the chain rule, the derivative of tan(3x) is 3sec^{2}(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 = 3sec^{2}(3x)

tan 3 x

► Derivative of tan 3 x = 3sec^{2}(3x)

tan 3x

► Derivative of tan 3x = 3sec^{2}(3x)

tan (3x)

► Derivative of tan (3x) = 3sec^{2}(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) = 3sec^{2}(3x). So to find the second derivative of tan(3x), we just need to differentiate 3sec^{2}(3x).

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

► The second derivative of tan(2x) is 18sec^{2}(3x)tan(3x)

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

We know how to differentiate x^{3} (the answer is 3x^{2})

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 ln^{3}x contains no parenthesis, we can still view it as a composite function (a function of a function).

We can write ln^{3}x as (ln(x))^{3}.

Now the function is in the form of x^{3}, 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) = x^{3} and the function g(x) = ln(x), then the function (ln(x))^{3} can be written as a composite function.

f(x) = x^{3}

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

= (3ln^{2}(x)).(1/x))

f(g(x)) = (ln(x))^{3}⇒f'(g(x)) = 3ln^{2}(x)

= 3ln^{2}(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.

ln^{3}x

► Derivative of ln^{3}x = 3ln^{2}(x)/x

ln^3x

► Derivative of ln^3x = 3ln^{2}(x)/x

ln 3 x

► Derivative of ln 3 x = 3ln^{2}(x)/x

(lnx)^3

► Derivative of (lnx)^3 = 3ln^{2}(x)/x

ln cubed x

► Derivative of ln cubed x = 3ln^{2}(x)/x

lnx3

► Derivative of lnx3 = 3ln^{2}(x)/x

ln^3

► Derivative of ln^3 = 3ln^{2}(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 = 3ln^{2}(x)/x. So to find the second derivative of ln^3x, we just need to differentiate 3ln^{2}(x)/x

We can use a combination of the chain rule and the quotient rule to find the derivative of 3ln^{2}(x)/x.

We can set f(x) = ln^{2}(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(-ln^{2}(x) + 2ln(x))/x^{2}

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

Note that in this post we will be looking at differentiating cot(2x) which is not the same as differentiating cot^{2}(x). Here is our post dealing with how to differentiate cot^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 cot(x) (the answer is cosec^{2}(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 -csc^{2}(x) The derivative of cot(z) with respect to z is -csc^{2}(z)

In a similar way, the derivative of cot(2x) with respect to 2x is -csc^{2}(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

= (-csc^{2}(2x)).(2)

f(g(x)) = cot(2x) ⇒f'(g(x)) = -cosec^{2}(2x)

= -2csc^{2}(2x)

Using the chain rule, the derivative of cot(2x) is -2csc^{2}(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 = -2csc^{2}(2x)

cot 2 x

► Derivative of cot 2 x = -2csc^{2}(2x)

cot 2x

► Derivative of cot 2x = -2csc^{2}(2x)

cot (2x)

► Derivative of cot (2x) = -2csc^{2}(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) = -2csc^{2}(2x). So to find the second derivative of cot(2x), we just need to differentiate -2csc^{2}(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 -2csc^{2}(x)cot(x)) and it gives us a result of 8csc^{2}(2x)cot(2x)

► The second derivative of cot(2x) is 8csc^{2}(2x)cot(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 ln^{2}(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) = ln^{2}(x)

Then remember that ln^{2}(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

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 ln(x) (the answer is l/x)

We know how to differentiate x^{2} (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 ln^{2}x contains no parenthesis, we can still view it as a composite function (a function of a function).

We can write ln^{2}x as (ln(x))^{2}.

Now the function is in the form of x^{2}, 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) = x^{2} and the function g(x) = ln(x), then the function (ln(x))^{2} can be written as a composite function.

f(x) = x^{2}

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.

ln^{2}x

► Derivative of ln^{2}x = 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

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 -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}

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^-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}.

Note that in this post we will be looking at differentiating csc(2x) which is not the same as differentiating csc^{2}(x). Here is our post dealing with how to differentiate csc^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 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 4csc^{3}(2x) + 4cot^{2}(2x)csc(2x)

► The second derivative of csc(2x) is 4csc^{3}(2x) + 4cot^{2}(2x)csc(2x)