Author Topic: Failed derivative (Read 558 times) Tweet Share

tony3272 · « **on:** May 24, 2011, 06:24:38 pm »

Hey guys. I got this question for our Differential Calculus assignment:
Find $\frac{\mathrm{d} y}{\mathrm{d} x}: \sin y=\cos ^2x$
So i'm just going along and i think "I cant be bothered with implicit diff, i'll just move the sin across".
$y= \sin^{-1}(\cos^2x)$
So just using the chain rule i get $\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2\cos x}{\sqrt {1+\cos ^2x}}$
However the actual answer is :
$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-\sin 2x}{\cos y}}$
This is the answer from just using implicit diff. I'm wondering as to why you get a different answer by using the inverse sine rather than implicit diff.

moekamo · « **Reply #1 on:** May 24, 2011, 07:04:12 pm »

they're the same thing:

$\begin{align} \frac{dy}{dx}&=\frac{-2\sin x}{\cos y} \\ &= \frac{-2\sin x \cos x}{\sqrt{1 - \sin ^2 y}} \\ &= \frac{-2 \sin x \cos x}{\sqrt{1-(\cos ^2 x)^2}} \\ &= \frac{-2 \sin x \cos x}{\sqrt{(1-\cos ^2 x)(1+ \cos ^2 x)}} \\ & = \frac{-2\sin x \cos x}{ \sin x \sqrt{1+\cos ^2 x}} \\ &= \frac{-2\cos x}{\sqrt{1+\cos ^2 x}} \end{align}$

so if they're the same, it shouldn't matter which one you did, unless the question specifically said use implicit differentiation...

tony3272 · « **Reply #2 on:** May 24, 2011, 07:23:47 pm »

Ohh lol, fail on my behalf again. I forgot that $sin^2 y=(cos^2 x)^2$ from the original equation, which made it very difficult to change that y into an x. $:-\$

Search the forums now!

We have moved!

Author Topic: Failed derivative (Read 558 times) Tweet Share

tony3272

Failed derivative

moekamo

Re: Failed derivative

tony3272

Re: Failed derivative

Recent Posts

User:
Password: