Question

Differentiate the function.

20.\(y = \ln \left( {\csc x - \cot x} \right)\)

Short answer

The derivative of the function is \(y' = \csc x\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunctionis shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Differentiate the function

Let \(u = \csc x - \cot x\) be the inner function.

Use the above formula to differentiate the function as shown below:

\(\begin{aligned}{c}y'&= \frac{d}{{dx}}\left( {\ln \left( {\csc x - \cot x} \right)} \right)\\&= \frac{1}{{\left( {\csc x - \cot x} \right)}} \cdot \frac{d}{{dx}}\left( {\csc x - \cot x} \right)\\&= \frac{1}{{\left( {\csc x - \cot x} \right)}} \cdot \left( { - \csc x\cot x + {{\csc }^2}x} \right)\\&= \frac{{\csc x\left( {\csc x - \cot x} \right)}}{{\csc x - \cot x}}\\&= \csc x\end{aligned}\)

Thus, the derivative of the function is \(y' = \csc x\).