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

2 Answers
Monzur R.
Apr 5, 2017

((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

Jim G.
Apr 5, 2017

((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

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