# Thread: Chain Rule - Book Example

1. ## Chain Rule - Book Example

I am having trouble understanding a portion of the textbook example. It's seems to be mostly with the simplifing of the answer.

Find dy/dx if y = u^3 - 3u^2 + 1 and u = x^2 + 2

dy/du = 3u^2 - 6u and du/dx = 2x

it follows that dx/dy = (3u^2 - 6u)(2x). *I'm fine so far, keep going*

We are substituting x^2 + 2 for u in the expression for dy/dx giving
[3(x^2+2)^2 - 6(x^2+2)](2x)

*I understand everything above but the next step confuses me*

6x(x^2+2)[(x^2+2)-2]
I don't know what steps they took to arrive at that equation. Could you possible show me the 'longer' process?

It goes on to give:
6x(x^2+2)(x^2)=
6x^3(x^2+2)

2. Originally Posted by becky
I am having trouble understanding a portion of the textbook example. It's seems to be mostly with the simplifing of the answer.

Find dy/dx if y = u^3 - 3u^2 + 1 and u = x^2 + 2

dy/du = 3u^2 - 6u and du/dx = 2x

it follows that dx/dy = (3u^2 - 6u)(2x). *I'm fine so far, keep going*

We are substituting x^2 + 2 for u in the expression for dy/dx giving
[3(x^2+2)^2 - 6(x^2+2)](2x)

*I understand everything above but the next step confuses me*

6x(x^2+2)[(x^2+2)-2]
I don't know what steps they took to arrive at that equation. Could you possible show me the 'longer' process?

It goes on to give:
6x(x^2+2)(x^2)=
6x^3(x^2+2)
First take common factor of 3 of the terms inside the square brackets outside
the square brackets

$\displaystyle [3(x^2+2)^2 - 6(x^2+2)](2x) =6x [(x^2+2)^2-2(x^2+2)]$

Now we also have a common factor of $\displaystyle (x^2+2)$ inside the square
brackets which we also take outside these brackets:

$\displaystyle [3(x^2+2)^2 - 6(x^2+2)](2x) =6x [(x^2+2)^2-2(x^2+2)]$$\displaystyle =6x (x^2+2)[(x^2+2)-2]$

RonL

3. Hello, Becky (and anyone listening in) . . .

You might as well get used to simplifying messy derivatives.
. . It comes up quite often.

Example: Find the derivative of $\displaystyle f(x)\:=\2x-3)^2(x+5)^4$

Product Rule: .$\displaystyle f'(x)\:=\2x-3)^2\cdot4(x+5)^3 + 2(2x-3)\cdot2\cdot(x+5)^4$

We have: .$\displaystyle f'(x) \:=\:4(2x-3)^2(x+5)^3 + 4(2x-3)(x+5)^4$ ...
You'd probably stop here, right? *

Look at the common factors: .$\displaystyle 4,\;(2x-3)\text{, and }(x+5)^3$

Factor them out: .$\displaystyle f'(x)\:=\:4(2x-3)(x+5)^3\,\left[(2x - 3) + (x + 5)\right]$

. . and we have: .$\displaystyle \boxed{f'(x)\:=\:4(2x-3)(x+5)^3(3x+2)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Your next question is (naturally) "Do we haaafta?"
(I'm not laughing at you . . . Years ago, I asked the same question.)

Normally, we take derivatives for a purpose, not just for exercise.

A common problem is to find the extreme values of a function.
We would set the derivative equal to zero and solve for $\displaystyle x.$

With your original form: .$\displaystyle 4(2x-3)^2(x+5)^3 + 4(2x-3)(x+5)^4 \:=\:0$
. . you might multiply it out and stare at the fifth-degree equation for days.

But recall the procedure for solving an equation of any degree:
. . Factor the expression, set each factor equal to zero, and solve.

With the factored form, we have: .$\displaystyle 4(2x-3)(x+5)^3(3x+2) \:=\:0$

. . and we can "eyeball" the solutions: .$\displaystyle x \:=\:\frac{3}{2},\;\text{-}5,\;\text{-}\frac{2}{3}$