Question

How do you differentiate `f(x)=(x-2)/sinx-cosx/(4-x)` using the quotient rule?

Answer 1

`d/dx[(x-2)/sinx-cosx/(4-x)]=(sinx-(x-2)cosx)/sin^2x+((4-x)(sinx)+cosx)/(4-x)^2`

Explanation:

The quotient rule states that
`d/dx(f(x)/g(x))=(g(x)*f'(x)-f(x)*g'(x))/((g(x))^2)`

Hence applying this rule to the given functions yields

`d/dx[(x-2)/sinx-cosx/(4-x)]=(sinx-(x-2)cosx)/sin^2x-((4-x)(-sinx)-cosx(-1))/(4-x)^2`