# How do you differentiate g(x)= 3tan4x *sin2x*cos2x using the product rule?

Nov 6, 2015

$\frac{12 \sin \left(2 x\right) \cos \left(2 x\right)}{{\cos}^{2} \left(4 x\right)} + 6 \tan \left(4 x\right) \cdot {\cos}^{2} \left(2 x\right) - 6 \tan \left(4 x\right) {\sin}^{2} \left(2 x\right)$

#### Explanation:

The rule is very simple, you derive one function and leave the others untouched, and then sum all these terms. So:

• Derive the first function and leave the other two untouched:

$\left(\frac{d}{\mathrm{dx}} 3 \tan \left(4 x\right)\right) \cdot \sin \left(2 x\right) \cdot \cos \left(2 x\right)$

$= \left(3 \cdot \frac{1}{{\cos}^{2} \left(4 x\right)} \cdot 4\right) \cdot \sin \left(2 x\right) \cdot \cos \left(2 x\right)$

• Derive the second function and leave the other two untouched:

$3 \tan \left(4 x\right) \cdot \frac{d}{\mathrm{dx}} \sin \left(2 x\right) \cdot \cos \left(2 x\right)$

$= 3 \tan \left(4 x\right) \cdot \left(\cos \left(2 x\right) \cdot 2\right) \cdot \cos \left(2 x\right)$

• Derive the third function and leave the other two untouched:

$3 \tan \left(4 x\right) \cdot \sin \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \cos \left(2 x\right)$

$= 3 \tan \left(4 x\right) \cdot \sin \left(2 x\right) \cdot \left(- \sin \left(2 x\right) \cdot 2\right)$

We can readjust the three terms:

1. $\left(3 \cdot \frac{1}{{\cos}^{2} \left(4 x\right)} \cdot 4\right) \cdot \sin \left(2 x\right) \cdot \cos \left(2 x\right) = \frac{12 \sin \left(2 x\right) \cos \left(2 x\right)}{{\cos}^{2} \left(4 x\right)}$

2. $3 \tan \left(4 x\right) \cdot \left(\cos \left(2 x\right) \cdot 2\right) \cdot \cos \left(2 x\right) = 6 \tan \left(4 x\right) \cdot {\cos}^{2} \left(2 x\right)$

3. $3 \tan \left(4 x\right) \cdot \sin \left(2 x\right) \cdot \left(- \sin \left(2 x\right) \cdot 2\right) = - 6 \tan \left(4 x\right) {\sin}^{2} \left(2 x\right)$

The result will be the sum of the three terms.