Question

How do you differentiate `f(x)= (2 x^2 + 7 x - 2)/ (x cos x )` using the quotient rule?

Answer 1

See below.

Explanation:

This will also require a use of the product rule for the function in the denominator.

The quotient rule states that for some function `h(x)` that can be expressed as `f(x)/g(x)`, `h'(x)` is given by `(f'(x)g(x)-f(x)g'(x))/(g(x))^2`

Using that idea we are going to need to find `f'(x)` and `g'(x)`:
`f'(x)` is just `4x+7` and `g'(x)` is `cosx-xsinx`

Substituting into the rule, the derivative is equal to:
`((4x+7)(xcosx)-(2x^2+7x-2)(cosx-xsinx))/(xcosx)^2`

You can then simplify as required.

Answer 2

`d/dx(f(x))=((2x^2+2)*cos x+(2x^2+7x-2)x*sin x)/(x*cos x)^2`

Explanation:

The given equation is

`f(x)=(2x^2+7x-2)/(x*cos x)`

We use the formula for derivative of rational expression

`d/dx(u/v)=(v*d/dx(u)-u*d/dx(v))/v^2`

`d/dx(f(x))=d/dx((2x^2+7x-2)/(x*cos x))=((x*cos x)*d/dx(2x^2+7x-2)-(2x^2+7x-2)*d/dx(x*cos x))/(x*cos x)^2`

`d/dx(f(x))=((x*cos x)*(4x+7)-(2x^2+7x-2)(-x*sin x+cos x))/(x*cos x)^2`

We simplify at this point

`d/dx(f(x))=((4x^2*cos x+7x*cos x)-(-2x^3*sin x-7x^2*sin x+2x*sin x+2x^2*cos x+7x*cos x-2*cos x))/(x*cos x)^2`

`d/dx(f(x))=(4x^2*cos x+7x*cos x+2x^3*sin x+7x^2*sin x-2x*sin x-2x^2*cos x-7x*cos x+2*cos x)/(x*cos x)^2`

`d/dx(f(x))=((2x^2+2)*cos x+(2x^2+7x-2)x*sin x)/(x*cos x)^2`

God bless....I hope the explanation is useful.