Differentiate a Function

**fuzzyx** · May 16th 2011, 11:26 PM

i got given this question for home work and i have know idea where to start

given that d sin(x)/dx = cos(x) use the chain rule to differentiate the function f(x)=+or-squre root of 1-sin squared(x) , pi/2<x<pi

help would be much appreciated

**amul28** · May 16th 2011, 11:53 PM

you know the identity

$sin^2x+cos^2x=1$

$\Rightarrow cos^2x=1-sin^2x$

$\Rightarrow cosx=\pm \sqrt{1-sin^2x}$

so your function will be $f(x)=cosx$

**bugatti79** · May 16th 2011, 11:54 PM

Originally Posted by fuzzyx

i got given this question for home work and i have know idea where to start

given that d sin(x)/dx = cos(x) use the chain rule to differentiate the function f(x)=+or-squre root of 1-sin squared(x) , pi/2<x<pi

help would be much appreciated

If you let u = 1-sin^2 (x) and use the chain rule in this format

$\displaystyle \frac{d \sqrt{u}}{du}\frac{du}{dx}$

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