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