# 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 $sin(x)$.

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

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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