# What is the derivative of sin^2(3x)/cos(2x)?

Apr 1, 2017

Derivative of $\frac{{\sin}^{2} 3 x}{\cos 2 x}$ is

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

#### Explanation:

In order to differentiate a function of a function, say $y , = f \left(g \left(x\right)\right)$, where we have to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we need to do (a) substitute $u = g \left(x\right)$, which gives us $y = f \left(u\right)$. Then we need to use a formula called Chain Rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$. In fact if we have something like $y = f \left(g \left(h \left(x\right)\right)\right)$, we can have $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{df}} \times \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dh}}$

We also use here quotient rule that is if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$

as $g \left(x\right) = {\sin}^{2} 3 x$, using chain rule $\frac{\mathrm{dg}}{\mathrm{dx}} = 2 \sin 3 x \times \cos 3 x \times 3$

= $6 \sin 3 x \cos 3 x$ and

as $h \left(x\right) = \cos \left(2 x\right)$ then $\frac{\mathrm{dh}}{\mathrm{dx}} = - \sin 2 x \times 2 = - 2 \sin 2 x$

Hence derivative of $\frac{{\sin}^{2} 3 x}{\cos 2 x}$ is

$\frac{6 \sin 3 x \cos 3 x \times \cos 2 x - \left(- 2 \sin 2 x\right) \times {\sin}^{2} 3 x}{{\cos}^{2} 2 x}$

= $\frac{6 \sin 3 x \cos 3 x \cos 2 x + 2 \sin 2 x {\sin}^{2} 3 x}{{\cos}^{2} 2 x}$

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

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

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