differentiation question.

[revinnova]

I have a little trouble understanding the circled problem. The part I'm stuck on is highlighted in green. How did the study guide get that?

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[soroban]

Hello, revinnova!

How did the study guide get that?

My queston: how come you didn't get that?
(It's part of the Quotient Rule.)

Differentiate: \(\displaystyle \L\, y\;=\;\frac{x}{a^2\sqrt{a^2\,-\,x^2}}\,+\,C\)

We have: \(\displaystyle \L\:y \;= \;\frac{1}{a^2}\,\cdot\,\frac{x}{(a^2\,-\,x^2)^{1/2}}\,+\,C\)

Then: \(\displaystyle \L\,\frac{dy}{dx}\;=\;\frac{1}{a^2}\,\cdot\,\frac{(a^2\,-\,x^2)^{\frac{1}{2}}\,\cdot\,1 \:-\:x\,\cdot\,\left[\overbrace{\frac{1}{2}(a^2\,-\,x^2)^{-\frac{1}{2}}(-2x)}\right] }{a^2\,-\,x^2}\)


Simplify the expression in brackets:

. . \(\displaystyle \L\frac{1}{2}(a^2\,-\,x^2)^{-\frac{1}{2}}(-2x) \;=\;\frac{1}{\not{2}}\,\cdot\,\frac{1}{(a^2\,-\,x^2)^{\frac{1}{2}}}\,\cdot(-\not{2}x) \;=\;\frac{-x}{\sqrt{a^2\,-\,x^2}}\)


Then we have: \(\displaystyle \L\:\frac{dy}{dx}\;=\;\frac{1}{a^2}\,\cdot\,\frac{\sqrt{a^2\,-\,x^2}\,-\,x\left(\frac{-x}{\sqrt{a^2\,-\,x^2}}\right) }{a^2\,-\,x^2}\) . . . Got it?

 

[revinnova]

Yeah, I guess I was too lazy to calculate it. Yeah, I got it. Thanks!