How do you find the derivative of sin^2(3x)/cos(2x)?

Nov 15, 2015

$f ' \left(x\right) = \frac{6 \sin \left(3 x\right) \cos \left(3 x\right) \cos \left(2 x\right) + 2 {\sin}^{2} \left(3 x\right) \sin \left(2 x\right)}{{\cos}^{2} \left(2 x\right)}$

Explanation:

Use the Quotient Rule.
$f ' \left(x\right) = \frac{\textcolor{red}{\frac{d}{\mathrm{dx}} \left[{\sin}^{2} \left(3 x\right)\right]} \cdot \cos \left(2 x\right) - {\sin}^{2} \left(3 x\right) \cdot \textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left[\cos \left(2 x\right)\right]}}{\cos} ^ 2 \left(2 x\right)$

Use the Chain Rule to find both derivatives inside the quotient rule.
color(red)(d/(dx)[sin^2(3x)])=2sin(3x)*d/(dx)[sin(3x)]=2sin(3x)*cos(3x)*d/(dx)[3x]=color(red)(6sin(3x)cos(3x)
color(blue)(d/(dx)[cos(2x)])=-sin(2x)*d/(dx)[2x]=color(blue)(-2sin(2x)

Plug back in.
$\textcolor{g r e e n}{f ' \left(x\right) = \frac{6 \sin \left(3 x\right) \cos \left(3 x\right) \cos \left(2 x\right) + 2 {\sin}^{2} \left(3 x\right) \sin \left(2 x\right)}{{\cos}^{2} \left(2 x\right)}}$