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 $$y=8\sin ^{1}(\frac{x}{4})\frac{x\sqrt{16x^2}}{2}$$ Can anyone give me some hints or guides how to differentiate it? Thanks. Because what I got from my working is $$\frac{2}{\sqrt{1\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16x^2}}\sqrt{16x^2})$$. The given answer in book is $$\frac{x^2}{\sqrt{16x^2}}$$ 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?

March 8th, 2015, 08:43 AM  #3  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,995 Thanks: 132 Math Focus: Trigonometry  Quote:
$\displaystyle =\frac{2}{\sqrt{\frac{16}{16}\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16x^2}}\frac{\sqrt{16x^2}^2}{\sqrt{16x^2}})$ $\displaystyle =\frac{2}{\sqrt{\frac{1}{16}(16x^2)}}+\frac{1}{2}(\frac{x^2}{\sqrt{16x^2}}\frac{16x^2}{\sqrt{16x^2}})$ $\displaystyle =\frac{2}{\sqrt{\frac{1}{16}}{\sqrt{16x^2}}}+\frac{1}{2}(\frac{x^2(16x^2)}{\sqrt{16x^2}})$ $\displaystyle =\frac{2}{\frac{1}{4}\sqrt{16x^2}}+\frac{1}{2}(\frac{x^216+x^2)}{\sqrt{16x^2}})$ $\displaystyle =\frac{2\cdot4}{\sqrt{16x^2}}+\frac{1}{2}(\frac{2x^216)}{\sqrt{16x^2}})$ $\displaystyle =\frac{8}{\sqrt{16x^2}}+\frac{1}{\cancel{2}2}(\frac{\cancel{2}(x^28)}{\sqrt{16x^2}})$ $\displaystyle =\frac{\cancel{8}+x^2\cancel{8}}{\sqrt{16x^2}}$ $\displaystyle =\frac{x^2}{\sqrt{16x^2}}$  
March 8th, 2015, 10:46 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,178 Thanks: 1646  Let $\displaystyle u = \sin^{1}\left(\frac{x}{4}\right)$, so that $x = 4\sin(u)$ and $\displaystyle \frac{dx}{du} = 4\cos(u) = \sqrt{16  x^2}\!$. $\displaystyle y = 8u  8\sin(u)\cos(u)$ $\displaystyle \frac{dy}{dx} = \frac{dy}{du}{\LARGE/}\frac{dx}{du} = \frac{8 + 8\sin^2(u)  8\cos^2(u)}{4\cos(u)} = \frac{16\sin^2(u)}{4\cos(u)} = \frac{x^2}{\sqrt{16  x^2}}$ 

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differentiation, inverse, sine 
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