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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: tony3272 on May 24, 2011, 06:24:38 pm

Title: Failed derivative
Post by: 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.
Title: Re: Failed derivative
Post by: moekamo 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...
Title: Re: Failed derivative
Post by: tony3272 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. :-\