# 7. Differentiating Powers of a Function

by M. Bourne

## Function of a Function

If *y* is a function of *u*, and *u* is a
function of *x*, then we say

"

yis a function of the functionu".

### Example 1

### Don't miss...

In this section:

Chain Rule

Power Rule

Consider the function

y= (5x+ 7)^{12}.

If we let *u* = 5*x* + 7 (the inner-most expression), then we could write our original function as

y=u^{12}

We have written *y* as a function of *u*, and in turn, *u* is a function of *x*.

This is a vital concept in differentiation, since many of the functions we meet from now on will be functions of functions, and we need to recognise them in order to differentiate them properly.

## Chain Rule

To find the derivative of a function of a function, we need to use the Chain Rule:

`(dy)/(dx) = (dy)/(du) (du)/(dx)`

This means we need to

- Recognise `u`
- Then we need to re-express `y`
- Then we differentiate `y` (with respect to `u`), then we re-express everything in terms of `x`.
- The next step is to find `(du)/dx`.
- Then we multiply `dy/(du)` and `(du)/dx`
*.*

### Answer for Example 1

In Example 1,

y=u^{12}, so `(dy)/(du) = 12u^11 = 12(5x+7)^11`

and

u= 5x+ 7, so `(du)/(dx) = 5`

So

`dy/dx=(dy)/(du) (du)/(dx) ` `= (12(5x+7)^11)(5) ` `= 60(5x+7)^11`

### Example 2

Find `dy/dx` if `y = (x^2+ 3)^5`.

Answer

In this case, we let *u* = *x*^{2} + 3 and
then *y* = *u*^{5}.

We see that:

*u*is a function of*x*and*y*is a function of*u.*

For the **chain rule**, we firstly need to find `(dy)/(du)` and `(du)/(dx)`:

`(dy)/(du)=5u^4=5(x^2+3)^4`; and

`(du)/(dx)=2x`

So

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=5(x^2+3)^4(2x)`

`=10x(x^2+3)^4`

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Play with the graph of this example on the Differentiation interactive applet page and explore what it means.

### Example 3

Find `dy/dx` if `y=sqrt(4x^2-x)`.

Answer

In this case, we let *u* = 4*x*^{2}
− *x * and then `y=sqrtu=u^(1/2)`.

Once again,

*u*is a function of*x*and*y*is a function of*u.*

Using the **chain rule**, we firstly need to find:

`(dy)/(du)=1/2u^(-1/2)=1/(2sqrtu)=1/(2sqrt(4x^2-x)`

and

`(du)/(dx)=8x-1`

So

`(dy)/(dx)=(dy)/(du)(du)/(dx)`

`=1/(2sqrt(4x^2-x))(8x-1)`

`=(8x-1)/(2sqrt(4x^2-x))`

**Note:** How does `u^(-1//2)=1/sqrtu`? See:

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You can play with this example on the Differentiation interactive applet page.

## The Derivative of a Power of a Function (Power Rule)

An extension of the **chain rule** is the **Power Rule** for differentiating. We are finding the **derivative** of
*u*^{n} (a power of a function):

`d/dxu^n=n u^(n-1)(du)/dx`

### Example 4

In the case of `y=(2x^3-1)^4` we have a
**power of a function.**

Answer

If we let *u *= 2*x*^{3} - 1 then *y *= *
u*^{4}.

So now

*y*is written as a power of*u*; and*u*is a function of*x*[*u*=*f*(*x*) ].

To find the derivative of such an expression, we can use our new rule:

`d/(dx)u^n=n u^(n-1)(du)/(dx`

where *u *= 2*x*^{3} − 1 and *n* =
4.

So

`(dy)/(dx)=n u^(n-1)(du)/(dx)`

`=[4(2x^3-1)^3][6x^2]`

`=24x^2(2x^3-1)^3`

We could, of course, use the **chain rule**, as
before**:**

`(dy)/(dx)=(dy)/(du)(du)/(dx`

Play with this example on the Differentiation interactive applet page.

### Example 5

If `y=1/x`, find `dy/dx`

Answer

We can write `y=1/x` as `y=x^-1`. So:

`dy/dx=(-1)x^-2 = -1/x^2`

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You can explore this example on the Differentiation interactive applet page.

## CHALLENGE

Find the derivative of `y=(x^2(3x+1))/(x^4+2)`

Answer

In this example, we have a **quotient**, where the
numerator is a **product***.*

Once again, we let `y=u/v`, where `u=x^2(3x+1)` and `v=x^4+2`.

The quotient formula requires `(du)/(dx)` but *u* is a product.

Let `u = pq` where *p* = *x*^{2} and
*q* = 3*x* + 1.

`(du)/(dx)=p(dq)/(dx)+q(dp)/(dx)`

`=(x^2)(3)+(3x+1)(2x)`

`=3x^2+6x^2+2x`

`=9x^2+2x`

We also require

`(dv)/(dx)=4x^3`

So

` (dy)/(dx)=(d(u/v))/(dx)=(v(du)/(dx)-u(dv)/(dx))/(v^2)`

`=((x^4+2)(9x^2+2x)-(x^2)(3x+1)(4x^3))/((x^4+2)^2)`

`=(9x^6+2x^5+18x^2+4x-12x^6-4x^5)/((x^4+2)^2)`

`=(-3x^6-2x^5+18x^2+4x)/((x^4+2)^2)`

Easy to understand math videos:

MathTutorDVD.com

Play with this challenge example on the Differentiation interactive applet page.

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