How do you find the derivative of y=ln(sin(x)) ?

May 19, 2018

$\cot x$

Explanation:

We use the chain rule, which states that,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = \sin x , \therefore \frac{\mathrm{du}}{\mathrm{dx}} = \cos x$.

Then, $y = \ln u , \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$.

Combining, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \cdot \cos x$

$= \cos \frac{x}{u}$

Substituting back $u = \sin x$, we get:

$= \cos \frac{x}{\sin} x$

Notice how it equals to:

$= {\left(\sin \frac{x}{\cos} x\right)}^{-} 1$

But $\sin \frac{x}{\cos} x = \tan x$, so we get:

$= {\left(\tan x\right)}^{-} 1$

$= \frac{1}{\tan} x$

$= \cot x$

May 19, 2018

"dy/dx=cotx

Explanation:

$\text{differentiate using the "color(blue)"chain rule}$

$\text{given "y=f(g(x))" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$y = \ln \left(\sin x\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sin} x \times \frac{d}{\mathrm{dx}} \left(\sin x\right)$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = \cos \frac{x}{\sin} x = \cot x$