My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By fysmat
Reply
 
LinkBack Thread Tools Display Modes
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.
jiasyuen is offline  
 
March 8th, 2015, 04:54 AM   #2
Senior Member
 
fysmat's Avatar
 
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
fysmat is offline  
March 8th, 2015, 08:43 AM   #3
Senior Member
 
Monox D. I-Fly's Avatar
 
Joined: Nov 2010
From: Indonesia

Posts: 2,000
Thanks: 132

Math Focus: Trigonometry
Quote:
Originally Posted by jiasyuen View Post
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.
$\displaystyle \frac{2}{\sqrt{1-\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\sqrt{16-x^2})$
$\displaystyle =\frac{2}{\sqrt{\frac{16}{16}-\frac{x^2}{16}}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\frac{\sqrt{16-x^2}^2}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2}{\sqrt{\frac{1}{16}(16-x^2)}}+\frac{1}{2}(\frac{x^2}{\sqrt{16-x^2}}-\frac{16-x^2}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2}{\sqrt{\frac{1}{16}}{\sqrt{16-x^2}}}+\frac{1}{2}(\frac{x^2-(16-x^2)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2}{\frac{1}{4}\sqrt{16-x^2}}+\frac{1}{2}(\frac{x^2-16+x^2)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{2\cdot4}{\sqrt{16-x^2}}+\frac{1}{2}(\frac{2x^2-16)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{8}{\sqrt{16-x^2}}+\frac{1}{\cancel{2}2}(\frac{\cancel{2}(x^2-8)}{\sqrt{16-x^2}})$
$\displaystyle =\frac{\cancel{8}+x^2\cancel{-8}}{\sqrt{16-x^2}}$
$\displaystyle =\frac{x^2}{\sqrt{16-x^2}}$
Monox D. I-Fly is offline  
March 8th, 2015, 10:46 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 19,528
Thanks: 1750

Quote:
Originally Posted by jiasyuen View Post
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 $\displaystyle \frac{dx}{du} = 4\cos(u) = \sqrt{16 - x^2}\!$.

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

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}{\LARGE/}\frac{dx}{du} = \frac{8 + 8\sin^2(u) - 8\cos^2(u)}{4\cos(u)} = \frac{16\sin^2(u)}{4\cos(u)} = \frac{x^2}{\sqrt{16 - x^2}}$
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

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 08:35 PM
Differentiation involving inverse sine and radical joeljacks Calculus 1 October 15th, 2012 08:16 PM
Inverse of sine squared skarface Trigonometry 2 March 11th, 2012 05:59 AM
Differentiation with Inverse Trigs titans4ever0927 Calculus 3 February 20th, 2009 07:22 AM
inverse sine values karel747 Calculus 1 June 1st, 2008 09:54 AM





Copyright © 2018 My Math Forum. All rights reserved.