# How do you differentiate g(x) = (1 + sin(2x))(1 +cos^2(2x))  using the product rule?

Feb 21, 2016

$2 \setminus \cos \left(2 x\right) \left(\setminus {\cos}^{2} \left(2 x\right) - 2 \setminus \sin \left(2 x\right) \left(\setminus \sin \left(2 x\right) + 1\right) + 1\right)$

#### Explanation:

\frac{d}{dx}((1+\sin (2x))(1+\cos ^2(2x))

Applying product rule: ${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = 1 + \setminus \sin \left(2 x\right) , g = 1 + \setminus {\cos}^{2} \left(2 x\right)$

$= \setminus \frac{d}{\mathrm{dx}} \left(1 + \setminus \sin \left(2 x\right)\right) \left(1 + \setminus {\cos}^{2} \left(2 x\right)\right) + \setminus \frac{d}{\mathrm{dx}} \left(1 + \setminus {\cos}^{2} \setminus \left(2 x\right)\right) \left(1 + \setminus \sin \left(2 x\right)\right)$ ...... equation (i)

Here, $\frac{d}{\mathrm{dx}} \left(1 + \setminus \sin \left(2 x\right)\right)$= $2 \setminus \cos \left(2 x\right)$

{ Applying sum/difference rule; ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

$= \setminus \frac{d}{\mathrm{dx}} \left(1\right) + \setminus \frac{d}{\mathrm{dx}} \left(\setminus \sin \left(2 x\right)\right)$

$\setminus \frac{d}{\mathrm{dx}} \left(1\right) = 0$ and

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \sin \left(2 x\right)\right)$ = $2 \cos \left(2 x\right)$ by applying chain rule

$\setminus \frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \setminus \frac{\mathrm{df}}{\mathrm{du}} \setminus \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$ }

Again,

$\setminus \frac{d}{\mathrm{dx}} \left(1 + \setminus {\cos}^{2} \left(2 x\right)\right)$ = $- 4 \setminus \cos \left(2 x\right) \setminus \sin \left(2 x\right)$

{ Applying sum/difference rule; ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

$\setminus \frac{d}{\mathrm{dx}} \left(1\right) + \setminus \frac{d}{\mathrm{dx}} \left(\setminus {\cos}^{2} \left(2 x\right)\right)$

$\setminus \frac{d}{\mathrm{dx}} \left(1\right) = 0$ and $\setminus \frac{d}{\mathrm{dx}} \left(\setminus {\cos}^{2} \left(2 x\right)\right) = - 4 \setminus \cos \left(2 x\right) \setminus \sin \left(2 x\right)$}

Finally,Substituting in equation (i)

$= 2 \setminus \cos \left(2 x\right) \left(1 + \setminus {\cos}^{2} \left(2 x\right)\right) + \left(- 4 \setminus \cos \left(2 x\right) \setminus \sin \left(2 x\right)\right) \left(1 + \setminus \sin \left(2 x\right)\right)$

Simplifying it,we get

$2 \setminus \cos \left(2 x\right) \left(\setminus {\cos}^{2} \left(2 x\right) - 2 \setminus \sin \left(2 x\right) \left(\setminus \sin \left(2 x\right) + 1\right) + 1\right)$