# What is the slope of the line tangent to the graph of the function f(x)=ln(sin^2(x+3)) at the point where x=pi/3?

Mar 14, 2018

See below.

#### Explanation:

If:

$y = \ln x \iff {e}^{y} = x$

Using this definition with given function:

${e}^{y} = {\left(\sin \left(x + 3\right)\right)}^{2}$

Differentiating implicitly:

${e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\sin \left(x + 3\right)\right) \cdot \cos \left(x + 3\right)$

Dividing by ${e}^{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(\sin \left(x + 3\right)\right) \cdot \cos \left(x + 3\right)}{e} ^ y$

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

Cancelling common factors:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(\cancel{\sin \left(x + 3\right)}\right) \cdot \cos \left(x + 3\right)}{{\sin}^{\cancel{2}} \left(x + 3\right)}$

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

We now have the derivative and will therefore be able to calculate the gradient at $x = \frac{\pi}{3}$

Plugging in this value:

$\frac{2 \cos \left(\left(\frac{\pi}{3}\right) + 3\right)}{\sin \left(\left(\frac{\pi}{3}\right) + 3\right)} \approx 1.568914137$

This is the approximate equation of the line:

$y = \frac{15689}{10000} x - \frac{1061259119}{500000000}$

GRAPH: