How do you differentiate #g(x) = sqrt(x^2-1)cot(7-x)# using the product rule?

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Answer 1 · 2016-06-22T10:52:58

#frac{d}{dx}(sqrt{x^2-1}cot (7-x))=frac{xcot (-x+7)}{\sqrt{x^2-1}}+frac{sqrt{x^2-1}}{sin ^2(-x+7)}#

#frac{d}{dx}(sqrt{x^2-1}cot (7-x))#

Applying product rule, #(fcdot g)^'=f^'cdot g+fcdot g^'#

#f=sqrt{x^2-1},g=cot (7-x)#

#=frac{d}{dx}(sqrt{x^2-1})cot (7-x)+frac{d}{dx}(cot (7-x))sqrt{x^2-1}#

We know,
#frac{d}{dx}(sqrt{x^2-1})=frac{x}{sqrt{x^2-1}}#

#frac{d}{dx}(cot (7-x))=frac{1}{sin ^2(7-x)}#

[Applying chain rule #frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}#

let #7-x=u#

#=frac{d}{du}(cot (u))frac{d}{dx}(7-x)#

we know,#frac{d}{du}(cot (u))=-frac{1}{sin ^2(u)}#
and,#frac{d}{dx}(7-x)=-1#

so,#=frac{1}{sin ^2(7-x)}#]

Finally,
#=frac{x}{sqrt{x^2-1}}cot (7-x)+frac{1}{sin ^2(7-x)}sqrt{x^2-1}#

simplifying it,
#=frac{xcot (-x+7)}{\sqrt{x^2-1}}+frac{sqrt{x^2-1}}{sin ^2(-x+7)}#