Find the Derivative

**KonaBear** · October 3rd 2012, 08:11 PM

Just started derivatives in Math 262 and we are learning the chain rule.
Then I get this:

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

Her is what I'm attempting...

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

Is this even close?

Thanks for any help you might have.

**Prove It** · October 3rd 2012, 08:23 PM

Originally Posted by KonaBear

Just started derivatives in Math 262 and we are learning the chain rule.
Then I get this:

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

Her is what I'm attempting...

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

Is this even close?

Thanks for any help you might have.

First of all, the derivative of a sum is equal to the sum of the derivatives. So the derivative of the first 2x is 2. The hard part will be evaluating the derivative of $\displaystyle \begin{align*} \left[ 2x + \left( 2x + 1 \right) \right]^3 \end{align*}$ . It's a composition of functions, so the chain rule will need to be used. I always use Leibnitz notation for the chain rule, since it is easier. In this case you will need to do $\displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align*}$ .

So first, if we have $\displaystyle \begin{align*} y = \left[ 2x + \left( 2x + 1 \right)^3 \right]^3 \end{align*}$ , then we let $\displaystyle \begin{align*} u = 2x + \left( 2x + 1 \right)^3 \implies y = u^3 \end{align*}$ . Then $\displaystyle \begin{align*}\frac{dy}{du} = 3u^2 = 3 \left[ 2x + \left( 2x + 1 \right)^3 \right]^2 \end{align*}$ .

As for finding $\displaystyle \begin{align*} \frac{du}{dx} \end{align*}$ , we notice that $\displaystyle \begin{align*} u = 2x + \left( 2x + 1 \right)^3 \implies \frac{du}{dx} = 2 + \frac{d}{dx} \left[ \left( 2x + 1 \right)^3 \right] \end{align*}$ . Can you use the Chain Rule to evaluate this final derivative?

**MarkFL** · October 3rd 2012, 08:24 PM

Post edited to prevent devaluation of above post...

**KonaBear** · October 3rd 2012, 08:34 PM

Ok so the final answer would be

2 + 3[2x + (2x+1)^3]^2 (2 + 3(2x + 1)^2) ?

**Prove It** · October 3rd 2012, 09:04 PM

Originally Posted by KonaBear

Ok so the final answer would be

2 + 3[2x + (2x+1)^3]^2 (2 + 3(2x + 1)^2) ?

Not quite, it'll actually be $\displaystyle \begin{align*} 2 + 3\left[ 2x + \left( 2x + 1 \right)^3 \right]^2 \left[ 2 + \mathbf{6}\left( 2x + 1 \right)^2 \right] \end{align*}$ . Why?

**KonaBear** · October 3rd 2012, 10:37 PM

I wish I could see it, but I don't. No idea where that 6 comes from.

**Prove It** · October 3rd 2012, 10:42 PM

When you evalute $\displaystyle \begin{align*} \frac{d}{dx}\left[ \left(2x+ 1 \right)^3 \right] \end{align*}$ , the "inner" function is u = 2x + 1. What's its derivative?

The "outer" function is $\displaystyle \begin{align*} u^3 \end{align*}$ . What's its derivative?

What do you get when you multiply them together?

**KonaBear** · October 3rd 2012, 10:44 PM

Ohhhhh ok! Thank you so much!

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