What is the derivative of # f(x)=(cos x*cscx) / (1+ tan^2 x) #?

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Answer 1 · 2015-11-18T09:02:57

#f'(x)=-2cos^2(x)-cot^2(x)#

First, simplify the expression by converting everything to #sin# and #cos#.

#f(x)=frac{cos(x)*frac{1}{sin(x)}}{frac{cos^2(x)}{cos^2(x)}+frac{sin^2(x)}{cos^2(x)}}#

#=frac{cos^3(x)}{sin(x)}#

#f'(x)=frac{d}{dx}(frac{cos^3(x)}{sin(x)})#

#=frac{sin(x)frac{d}{dx}(cos^3(x))-cos^3(x)frac{d}{dx}(sin(x))}{sin(x)^2}#

#=frac{sin(x)(3cos^2(x)(-sin(x)))-cos^3(x)(cos(x))}{sin^2(x)}#

#=frac{-3sin^2(x)cos^2(x)-cos^4(x)}{sin^2(x)}#

#=frac{-3sin^2(x)cos^2(x)+cos^2(x)(sin^2(x)-1)}{sin^2(x)}#

#=-3cos^2(x)+cos^2(x)-cot^2(x)#

#=-2cos^2(x)-cot^2(x)#