Chain rule

Summary and examples

When we want to differentiate a function we can use the table of derivatives of standard functions, sometimes in combination with the product and quotient rule. Sometimes this is not possible, see the following example:

y=(5x^2+3x+1)^{10}

This function can not be differentiated easily by merely using the table and product or quotient rules. In such cases we can use the chain rule.

There are both a formal and informal version of the chain rule. We explain the formal way by the following example.

Example 1

Differentiate:

y=(5x^2+3x+1)^{10}

This function is not in the table of derivatives of standard functions and the product rule is not applicable.

We apply the transformation:

u=5x^2+3x+1

that is to say, we switch to another variable. Then we get:

y=u^{10}

Now we apply the following rule (the chain rule):

\displaystyle\frac{dy}{dx}=\displaystyle\frac{dy}{du}\displaystyle\frac{du}{dx}

and get:

\displaystyle\frac{dy}{du}=10u^9

and:

\displaystyle\frac{du}{dx}=10x+3

And thus:

\displaystyle\frac{dy}{dx}=y'=10u^9(10x+3)

We want the result in only the variable x and get:

y'=10(5x^2+3x+1)^9(10x+3)

Example 2

Differentiate:

y=\ln(x^2+1)

If we define:

u=x^2+1

we get:

y=\ln(u)

and this function is in our table with standard functions. Applying the chain rule yields:

y'=\displaystyle\frac{1}{u}\cdot2x

and, back to x:

y'=\displaystyle\frac{1}{x^2+1}\cdot2x=\displaystyle\frac{2x}{x^2+1}

The informal method works as follows. We look again at the previous functions. If we want to differentiate:

y=(5x^2+3x+)^{10}

we could naively apply the table:

y=x^n\longrightarrow y'=nx^{n-1}

which would have resulted in:

y'=10(5x^2+3x+1)^9

but this is wrong. To get the correct answer we have to multiply this preliminary result by the derivative of the function that took the place of x in the function y=x^n. In this case we have to multiply by the derivative of 5x^2+3x+1 which is 10x+3. Then the result is:

y=10(5x^2+3x+1)^9(10x+3)

In the second example the ‘naive differentiator’ would have got as the derivative:

y'=\displaystyle\frac{1}{x^2+1}

but this is not the right answer. Again, to get the correct derivative, we have to multiply this result by the derivative of x^2+1 which is 2x. The correct answer is:

y'=\displaystyle\frac{2x}{x^2+1}

A few more examples.

Example 3

y=\sin(3x+1)\Longrightarrow y'=\cos(3x+1)\cdot3=3\cos(3x+1)

In this example we differentiate the standard sin-function and multiply the result by the derivative of 3x+1.

Example 4

y=\sin(x^2)\Longrightarrow y'=\cos(x^2)\cdot2x=2x\cos(x^2)

Again the sin-function is the standard function and the derivative is cos(x). We have to multiply this result by the derivative of x^2.

Example 5

Differentiate:

y=\sin^2(x)

In this example we need to differentiate a square of the function sin(x). This means that x^n is the standard function and that sin(x) is in place of x.

First we differentiate:

y=[\sin(x)]^2

and multiply the result by the derivative of sin(x).

The correct answer is thus:

y'=2\sin(x)\cos(x)

Note that the chain rule is an extra method to existing methods: the derivatives of standard functions and the product and quotient rule. We can combine these methods to determine a derivative, see the following example.

Example 6

Differentiate:

y=\displaystyle\frac{\sin(2x)}{x}

In this case the numerator is:

f(x)=\sin(2x)

and the denominator is:

g(x)=x

and thus we have:

f'(x)=\cos(2x)\cdot2=2\cos(2x) (chain rule)

g'(x)=1 (table)

and thus the result is:

y'=\displaystyle\frac{2x\cos(2x)-\sin(2x)}{x^2}

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