March 8th, 2015, 03:05 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, 05:54 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, 09:43 AM  #3  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 2,000 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, 11:46 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,951 Thanks: 1842  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 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Differentiation of sine function w/ arctan argument  henoshaile  Calculus  5  October 15th, 2012 09:35 PM 
Differentiation involving inverse sine and radical  joeljacks  Calculus  1  October 15th, 2012 09:16 PM 
Inverse of sine squared  skarface  Trigonometry  2  March 11th, 2012 06:59 AM 
Differentiation with Inverse Trigs  titans4ever0927  Calculus  3  February 20th, 2009 08:22 AM 
inverse sine values  karel747  Calculus  1  June 1st, 2008 10:54 AM 