Differentiating Big Multiple

• Oct 14th 2009, 07:33 AM
Kataangel
Differentiating Big Multiple
Hi I need urgent help for this expression as I have no idea how to get its derivative:

$\displaystyle f(x)=(x^2+1)^100(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

I have to use the ln function but I can only come up to here:

$\displaystyle \frac{d}{dx}ln(f(x))=100\frac{2x}{x^2+1}+2\frac{3x ^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$

Specifically I don't know how to get rid of the natural log function on the left side of the equation. I would appreciate if anyone could tell me how?
• Oct 14th 2009, 09:07 AM
ramiee2010
Quote:

Originally Posted by Kataangel
Hi I need urgent help for this expression as I have no idea how to get its derivative:

$\displaystyle f(x)=(x^2+1)^100(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

I have to use the ln function but I can only come up to here:

$\displaystyle \frac{d}{dx}ln(f(x))=100\frac{2x}{x^2+1}+2\frac{3x ^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$

Specifically I don't know how to get rid of the natural log function on the left side of the equation. I would appreciate if anyone could tell me how?

$\displaystyle \frac{d}{dx}ln(f(x))= \frac{1}{f(x)}f'(x)$
$\displaystyle therefore \quad \frac{1}{f(x)}f'(x)==\frac{200x}{x^2+1}+\frac{6x^2 }{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$
$\displaystyle therefore \quad f'(x)=f(x)\times (\frac{200x}{x^2+1}+\frac{6x^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3})$
• Oct 14th 2009, 09:19 AM
Soroban
Hello, Kataangel!

Quote:

Hi, I need urgent help for this expression as I have no idea how to get its derivative.
. .
Why do you say that? You obviously know how to use logs and to differentiate!

$\displaystyle f(x)=(x^2+1)^{100}(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

I have to use the ln function but I can only come up to here:

$\displaystyle \frac{d}{dx}ln(f(x))=100\frac{2x}{x^2+1}+2\frac{3x ^2}{x^3-7}+\frac{1}{x-1}-\frac{10x^9+1}{x^{10}+x+3}$

Specifically I don't know how to get rid of the natural log function on the left side of the equation.
I would appreciate if anyone could tell me how.

Write it like this: .$\displaystyle {\color{blue}y} \;=\;(x^2+1)^{100}(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}$

Take logs: .$\displaystyle \ln(y) \;=\;100\ln(x^2+1) + 2\ln(x^3-7) + \ln(x-1) - \ln(x^{10}+x+3)$

Diffrerentiate: .$\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\frac{200x}{x^2+1} + \frac{6x^2}{x^3-7} + \frac{1}{x-1} - \frac{10x^9 + 1}{x^{10}+x+3}$

Then: .$\displaystyle \frac{dy}{dx} \;=\;y\bigg[\frac{200x}{x^2+1} + \frac{6x^2}{x^3-7} + \frac{1}{x-1} - \frac{10x^9 + 1}{x^{10}+x+3}\bigg]$

. . . . . $\displaystyle \frac {dy}{dx}\;=\;(x^2+1)^{100}(x^3-7)^2(x-1)(x^{10}+x+3)^{-1}\bigg[\frac{200x}{x^2+1} + \frac{6x^2}{x^3-7} + \frac{1}{x-1} - \frac{10x^9 + 1}{x^{10}+x+3}\bigg]$