1. chain rule + differentiation

missed a calc lesson and trying to play catch up. a little confused on this one:

find the derivative of the function using chain rule or logarithmic differentiation

f(x) = (x^2 + 2)^200 * (x^3 - x)^10

any step by step help is greatly appreciated

2. Originally Posted by JGaraffa
missed a calc lesson and trying to play catch up. a little confused on this one:

find the derivative of the function using chain rule or logarithmic differentiation

f(x) = (x^2 + 2)^200 * (x^3 - x)^10

any step by step help is greatly appreciated
Two solve this I'd use two facts.

If $\displaystyle u = u(x)$ and $\displaystyle v = v(x)$, then $\displaystyle \frac{d}{dx}(uv) = v.du + u.dv$, (product rule)

And the for the chain rule.

$\displaystyle \frac{d}{dx}(f(x))^n = n.f'(x).(f(x))^{n-1}$.

This is all you need to solve your question. Post all your working and we can help if you get stuck.

3. Hello, JGaraffa!

I'll do it both ways for you . . .

Differentiate using the chain rule or log differentiation:

. . $\displaystyle f(x) \:=\:(x^2 + 2)^{200}(x^3 - x)^{10}$

Product and Chain Rules:

. . $\displaystyle f'(x) \;=\;(x^2+2)^{200}\cdot10(x^3-x)^0(3x^2-1) + (x^3-x)^{10}\cdot200(x^2+2)^{199}2x$

Factor: .$\displaystyle f'(x) \;=\;10(x^2+2)^{199}(x^3-x)^9\,\bigg[(x^2+2)(3x^2-1) + (40x(x^3-x)\bigg]$

. . . . . . . . . $\displaystyle =\;10(x^2+2)^{199}(x^3-x)^9\bigg[3x^4 + 6x^2 - x^2 - 2 + 40x^4 - 40x^2\bigg]$

. . . . . . . . . $\displaystyle =\;10(x^2+2)^{199}(x^3-x)^9\left(43x^4 - 35x^2 - 2\right]$

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

Logarithmic Differentiaton:

Take logs:

. . $\displaystyle \ln(y) \;=\;\ln\bigg[(x^2+2)^{200}(x^3-x)^{10}\bigg]$

. . . . . .$\displaystyle =\;\ln(x^2+2)^{200} + \ln(x^3-x)^{10}$

. . . . . .$\displaystyle =\;200\ln(x^2+2) + 10\ln(x^3-x)$

Differentiate implicitly:

. . $\displaystyle \frac{1}{y}\cdot y' \;=\;200\cdot\frac{2x}{x^2+2} + 10\cdot\frac{3x^2-1}{x^3-x}$

. . . . $\displaystyle \frac{y'}{y} \;=\;10\bigg[\frac{40x}{x^2+2} + \frac{3x^2-1}{x^3-x}\bigg]$

. . . . $\displaystyle \frac{y'}{y} \;=\;10\cdot \frac{43x^4 - 35x^2 - 2}{(x^2+2)(x^3-x)}$

. . . . $\displaystyle y' \;=\;y\cdot10\cdot\frac{43x^4 - 35x^2 - 2}{(x^2+2)(x^3-x)}$

. . . . $\displaystyle y'\;=\;(x^2+2)^{200}(x^3-x)^{10}\cdot 10 \cdot \frac{43x^4-35x^2-2}{(x^2+2)(x^3-x)}$

. . . . $\displaystyle y' \;=\;10(x^2+2)^{199}(x^3-x)^9(43x^4-35x^2-2)$

i set mine up a little differently, however

i used (x^2 + 2)^200 as f(x)

and (x^3 - x)^10 as g(x)

so when i did chain rule it came out as

f(x) = 2x[200(x^2 + 2)^199]
g(x) = 3x^2[10(x^3 - x)^9]

right so far?

then i used product rule and came up with:

2x[200(x^2 + 2)^199] * (x^3-x)^10 + (x^2)^200 * 3x^2[10(x^3 - x)^9]

right? a little confused how to simplify from here. can you help?

usually my prof lets us leave it like this but i may as well leave that as well as the way you simplified it, just in case.

thanks again!