Question

How do you differentiate `f(x)= (9-x)(5/(x^2 -4))` using the product rule?

Answer 1

`f'(x)=5*(x^2-18x+4)/((x-2)^2*(x+2)^2)`

Explanation:

After the product rule
`(uv)'=u'v+uv'`
we get

`f'(x)=-5/(x^2-4)+(9-x)*(-5)*(x^2-4)^(-2)*2*x`

which simplifies to
`f'(x)=5*(x^2-18x+4)/((x-2)^2*(x+2)^2)`

Answer 2

`f'(x)=(5(x^2 -18x+4))/(x^2 -4)^2`

Explanation:

`f(x)= (9-x)(5/(x^2 -4))`

Apply product rule,

`f'(x)= [d/dx(9-x)] (5/(x^2 -4))+[d/dx(5/(x^2 -4))] (9-x)`

Differentiate inner terms,

`f'(x)=(-1) (5/(x^2 -4))+(-(10x)/(x^2 -4)^2) (9-x)`

Simplify,

`f'(x)=-5/(x^2 -4)-(10x(9-x))/(x^2 -4)^2`

Common denominator,

`f'(x)=-(5(x^2 -4))/(x^2 -4)^2-(10x(9-x))/(x^2 -4)^2`

Simplify,

`f'(x)=-(5x^2 -20)/(x^2 -4)^2-(90x-10x^2)/(x^2 -4)^2`

Combine,

`f'(x)=(-5x^2 +20-90x+10x^2)/(x^2 -4)^2`

Simplify,

`f'(x)=(5x^2 -90x+20)/(x^2 -4)^2`

Factorise,

`f'(x)=(5(x^2 -18x+4))/(x^2 -4)^2`