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.

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- Oct 3rd 2012, 08:11 PMKonaBearFind the Derivative
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. - Oct 3rd 2012, 08:23 PMProve ItRe: Find the Derivative
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 $\displaystyle \displaystyle \begin{align*} \left[ 2x + \left( 2x + 1 \right) \right]^3 \end{align*}$. 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 $\displaystyle \displaystyle \begin{align*} \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \end{align*}$.

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

As for finding $\displaystyle \displaystyle \begin{align*} \frac{du}{dx} \end{align*}$, we notice that $\displaystyle \displaystyle \begin{align*} u = 2x + \left( 2x + 1 \right)^3 \implies \frac{du}{dx} = 2 + \frac{d}{dx} \left[ \left( 2x + 1 \right)^3 \right] \end{align*}$. Can you use the Chain Rule to evaluate this final derivative? - Oct 3rd 2012, 08:24 PMMarkFLRe: Find the Derivative
Post edited to prevent devaluation of above post...

- Oct 3rd 2012, 08:34 PMKonaBearRe: Find the Derivative
Ok so the final answer would be

2 + 3[2x + (2x+1)^3]^2 (2 + 3(2x + 1)^2) ? - Oct 3rd 2012, 09:04 PMProve ItRe: Find the Derivative
- Oct 3rd 2012, 10:37 PMKonaBearRe: Find the Derivative
I wish I could see it, but I don't. No idea where that 6 comes from.

- Oct 3rd 2012, 10:42 PMProve ItRe: Find the Derivative
When you evalute $\displaystyle \displaystyle \begin{align*} \frac{d}{dx}\left[ \left(2x+ 1 \right)^3 \right] \end{align*}$, the "inner" function is u = 2x + 1. What's its derivative?

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

What do you get when you multiply them together? - Oct 3rd 2012, 10:44 PMKonaBearRe: Find the Derivative
Ohhhhh ok! Thank you so much!