Question

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

Answer 1

`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)}`

Explanation:

`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)}`