Need Help Identifying Technique

**Algebrah** · October 14th 2011, 06:27 AM

Hi everyone! I just got out of class and was reviewing my notes. In one of the example exercises, I found a technique that I'm not familiar with. Here's the example question:

$f(x) = \frac{x^8 \cos ^3 x}{\sqrt{x-1}}$

$\ln y = \ln (\frac{x^8 \cos ^3 x}{(x-1)^\frac{1}{2}}})$

$\ln y = \ln (x^8 \cos ^3x) - \ln(x-1)^\frac{1}{2}$

$\ln y = \ln(x^8) + \ln (\cos ^3x) - \frac{1}{2} \ln(x-1)$

$\ln y = 8\ln x + 3\ln \cos x - \frac{1}{2} \ln(x-1)$

$\frac{1}{y} \frac{dy}{dx} = \frac{8}{x} + \frac{3}{\cos x} \sin x - \frac{1}{2} \frac{1}{(x-1)}$

$\frac{dy}{dx} = \frac{x^8 \cos ^3x}{\sqrt{x-1}} (\frac{8}{x} - \frac{3 \sin x}{\cos x} - \frac{1}{2(x-1)})$

The problem that I have is in the final step (this also isn't fully simplified, but I didn't have time to type the whole thing since I just taught myself LaTeX in the past 20 minutes

). Where does the

$\frac{x^8 \cos ^3x}{\sqrt{x-1}}$

come from? I heard the word "substitution" mentioned, but that's all I know. An explanation would be greatly appreciated! Also, would it be possible for you to show a minor amount of work to help explain the technique?

Thanks a lot!

**Deveno** · October 14th 2011, 06:38 AM

that's y!

in the step before, on the left, you had:

(1/y)(dy/dx) = stuff, so...

dy/dx = (y)(stuff)

**Algebrah** · October 14th 2011, 07:40 AM

Oh! That's so obvious!

Thanks a lot!

:manythanks:

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