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

Question

Scarlett Allison · Accepted Answer

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

Elijah Abram · Answer

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

William Abdalla · Answer

undefined To find the derivative of $ \frac{x^2+x}{\sin(x)} $, you can use the quotient rule, which states that for functions $ u(x) $ and $ v(x) $, the derivative of $ \frac{u(x)}{v(x)} $ is given by:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $$

Where $ u'(x) $ and $ v'(x) $ denote the derivatives of $ u(x) $ and $ v(x) $ respectively.

Let $ u(x) = x^2 + x $ and $ v(x) = \sin(x) $. Then:

$$ u'(x) = 2x + 1 $$
$$ v'(x) = \cos(x) $$

Substituting these into the quotient rule formula:

$$ \frac{d}{dx} \left( \frac{x^2+x}{\sin(x)} \right) = \frac{(2x + 1)\sin(x) - (x^2 + x)\cos(x)}{(\sin(x))^2} $$