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 Calculus Calculus Math Forum

 March 8th, 2015, 01:05 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Differentiation of inverse sine. I'm given a function $$y=8\sin ^{-1}(\frac{x}{4})-\frac{x\sqrt{16-x^2}}{2}$$ Can anyone give me some hints or guides how to differentiate it? Thanks. Because what I got from my working is $$\frac{2}{\sqrt{1-\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\sqrt{16-x^2})$$. The given answer in book is $$\frac{x^2}{\sqrt{16-x^2}}$$ It seems my answer is different with the given answer. Is there some short method to solve this? Thanks a lot. March 8th, 2015, 04:54 AM #2 Senior Member   Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Have you tried to simplify your result? Thanks from skeeter March 8th, 2015, 08:43 AM   #3
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 Originally Posted by jiasyuen I'm given a function $$y=8\sin ^{-1}(\frac{x}{4})-\frac{x\sqrt{16-x^2}}{2}$$ Can anyone give me some hints or guides how to differentiate it? Thanks. Because what I got from my working is $$\frac{2}{\sqrt{1-\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\sqrt{16-x^2})$$. The given answer in book is $$\frac{x^2}{\sqrt{16-x^2}}$$ It seems my answer is different with the given answer. Is there some short method to solve this? Thanks a lot.
$\displaystyle \frac{2}{\sqrt{1-\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\sqrt{16-x^2})$
$\displaystyle =\frac{2}{\sqrt{\frac{16}{16}-\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\frac{\sqrt{16-x^2}^2}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2}{\sqrt{\frac{1}{16}(16-x^2)}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\frac{16-x^2}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2}{\sqrt{\frac{1}{16}}{\sqrt{16-x^2}}}+\frac{1}{2}(\frac{x^2-(16-x^2)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2}{\frac{1}{4}\sqrt{16-x^2}}+\frac{1}{2}(\frac{x^2-16+x^2)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2\cdot4}{\sqrt{16-x^2}}+\frac{1}{2}(\frac{2x^2-16)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{8}{\sqrt{16-x^2}}+\frac{1}{\cancel{2}2}(\frac{\cancel{2}(x^2-8)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{\cancel{8}+x^2\cancel{-8}}{\sqrt{16-x^2}}$
$\displaystyle =\frac{x^2}{\sqrt{16-x^2}}$ March 8th, 2015, 10:46 AM   #4
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Quote:
 Originally Posted by jiasyuen Is there some short method to solve this?
Let $\displaystyle u = \sin^{-1}\left(\frac{x}{4}\right)$, so that $x = 4\sin(u)$ and $\displaystyle \frac{dx}{du} = 4\cos(u) = \sqrt{16 - x^2}\!$.

$\displaystyle y = 8u - 8\sin(u)\cos(u)$

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}{\LARGE/}\frac{dx}{du} = \frac{8 + 8\sin^2(u) - 8\cos^2(u)}{4\cos(u)} = \frac{16\sin^2(u)}{4\cos(u)} = \frac{x^2}{\sqrt{16 - x^2}}$ Tags differentiation, inverse, sine ### differentation my math forum

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