# What is the equation of the line normal to f(x)= xsin^2(2x)  at x=pi/8?

Jan 5, 2016

$y - \frac{\pi}{16} = \frac{2 + \pi}{4} \left(x - \frac{\pi}{8}\right)$

#### Explanation:

Use the product rule to find $f ' \left(x\right)$:

$f ' \left(x\right) = {\sin}^{2} \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(x\right) + x \frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(2 x\right)\right)$

Find each derivative individually.

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

The next will require the chain rule.

$\frac{d}{\mathrm{dx}} \left({\sin}^{2} \left(2 x\right)\right) = 2 \sin \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(2 x\right)\right)$

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

Plug these back in.

$f ' \left(x\right) = {\sin}^{2} \left(2 x\right) + 4 x \sin \left(2 x\right) \cos \left(2 x\right)$

Now, to find the slope of the tangent line at $x = \frac{\pi}{8}$, find $f ' \left(\frac{\pi}{8}\right)$.

$f ' \left(\frac{\pi}{8}\right) = {\sin}^{2} \left(\frac{\pi}{4}\right) + \frac{\pi}{2} \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)$

$= \frac{1}{2} + \frac{\pi}{2} \left(\frac{1}{2}\right) = \frac{2 + \pi}{4}$

To find the point the tangent line will intersect, find $f \left(\frac{\pi}{8}\right)$.

$f \left(\frac{\pi}{8}\right) = \frac{\pi}{8} {\sin}^{2} \left(\frac{\pi}{4}\right) = \frac{\pi}{8} \left(\frac{1}{2}\right) = \frac{\pi}{16}$

The tangent line will intersect the point $\left(\frac{\pi}{8} , \frac{\pi}{16}\right)$ and have a slope of $\frac{2 + \pi}{4}$.

Relate this information in a line in point-slope form:

$y - \frac{\pi}{16} = \frac{2 + \pi}{4} \left(x - \frac{\pi}{8}\right)$

graph{(y-pi/16-(2+pi)/4(x-pi/8))(x(sin(2x))^2-y)=0 [-1.495, 2.35, -0.546, 1.376]}

Graphed are the function and its tangent line.