frac{d}{dx}(sqrt{x^2-1}cot (7-x))ddx(√x2−1cot(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)}
