# Step-by-step Solution

## Derive the function $\ln\left(\frac{1+\cos\left(x\right)}{1-\cos\left(x\right)}\right)$ with respect to x

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### Videos

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

## Step-by-step explanation

Problem to solve:

$\frac{d}{dx}\left(\ln\left(\frac{1+\cos\left(x\right)}{1-\cos\left(x\right)}\right)\right)$
1

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{\frac{1+\cos\left(x\right)}{1-\cos\left(x\right)}}\cdot\frac{d}{dx}\left(\frac{1+\cos\left(x\right)}{1-\cos\left(x\right)}\right)$
2

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

$\frac{-\sin\left(x\right)\left(1-\cos\left(x\right)\right)-\sin\left(x\right)\left(1+\cos\left(x\right)\right)}{\left(1-\cos\left(x\right)\right)^2}\cdot\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}$
$\frac{d}{dx}\left(\ln\left(\frac{1+\cos\left(x\right)}{1-\cos\left(x\right)}\right)\right)$

### Main topic:

Differential calculus

~ 0.63 seconds

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