Question

How do you find the derivative of `4-(x^2)sinx`?

Answer 1

`-2xsin(x)-x^2cos(x)`

Explanation:

By using the product rule
`(f(x)*g(x))'=f(x)'*g(x)+f(x)*g(x)'`

`k(x)=4-x^2sin(x)`
`k'(x)=0-((x^2)' * sin(x)+x^2*(sin(x))')`
`k'(x)=-2xsin(x)-x^2cos(x)`