Question

How do you find the derivative of `(x^2+x)/sinx`?

Answer 1

`((x^2+x)/sinx)^'=((2x+1)sinx-(x^2+x)cosx)/sin^2x`

Explanation:

Use the quotient rule:

`((f(x))/(g(x)))^'=(g(x)f'(x)-f(x)g'(x))/(g(x))^2`

`((x^2+x)/sinx)^'=((2x+1)sinx-(x^2+x)cosx)/sin^2x`

Answer 2

`((2x+1)sinx-(x^2+x)cosx)/sin^2x`

Explanation:

differentiate using the `color(blue)"quotient rule"`

`"Given " f(x)=(g(x))/(h(x))" then"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))larr" quotient rule"`

`"here " g(x)=x^2+xrArrg'(x)=2x+1`

`"and " h(x)=sinxrArrh'(x)=cosx`

`rArrf'(x)=(sinx(2x+1)-(x^2+x)cosx)/sin^2x`