Question

How do you find the derivative of $\ln (\tan x)?$

Answer 1

Hint:The given function is a composite function that is a function that consists of another function or the argument of a function is another function. Composite functions can be differentiated with the help of chain rule, let us take an example of a composite function $f(x)=h(g(x))$, its derivative will be given as follows:
${\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{d\left(h(g(x)\right)}{d\left(g(x)\right)}}\times {\displaystyle \frac{d\left(g(x)\right)}{dx}}$

Formula used:
Chain rule: If a function $f(x)$ is composition of various function as follows$f(x)=g_1(g_2(g_3(....g_1(x))))$ then ${\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{d\left(g_1(g_2(g_3(....g_n(x))))\right)}{d\left(g_2(g_3(....g_n(x)))\right)}}\times {\displaystyle \frac{d\left(g_2(g_3(....g_n(x)))\right)}{d\left(g_3(....g_n(x))\right)}}\times {\displaystyle \frac{d\left(g_3(....g_n(x))\right)}{d\left((....g_n(x)\right)}}\times ......\times {\displaystyle \frac{d\left(g_n(x)\right)}{dx}}$
Derivative of logarithm function: ${\displaystyle \frac{d\ln x}{dx}}={\displaystyle \frac{1}{x}}$
And derivative of tangent function: ${\displaystyle \frac{d\tan x}{dx}}={\sec }^{2}x$

Complete step by step answer:
In order to find the derivative of the function $f(x)=\ln (\tan x)$, we have to use chain rule because this is a composite function, tangent function is the argument of the logarithmic function in this given composite function. Let us understand chain rule in order to solve this problem.If a function $f(x)$ is composition of various function as follows
$f(x)=g_1(g_2(g_3(....g_1(x))))$ then its derivative will be given as
${\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{d\left(g_1(g_2(g_3(....g_n(x))))\right)}{d\left(g_2(g_3(....g_n(x)))\right)}}\times {\displaystyle \frac{d\left(g_2(g_3(....g_n(x)))\right)}{d\left(g_3(....g_n(x))\right)}}\times {\displaystyle \frac{d\left(g_3(....g_n(x))\right)}{d\left((....g_n(x)\right)}}\times ......\times {\displaystyle \frac{d\left(g_n(x)\right)}{dx}}$
In the given function
$f(x)=\ln (\tan x)$
Taking derivative both sides with respect to $x$
${\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{d\left(\ln (\tan x)\right)}{dx}}$
Now applying chain rule to the given function $f(x)=\ln (\tan x)$, we will get
${\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{d\left(\ln (\tan x)\right)}{d(\tan x)}}\times {\displaystyle \frac{d\left(\tan x\right)}{dx}}$
We know that,
${\displaystyle \frac{d\ln x}{dx}}={\displaystyle \frac{1}{x}}\text{and}{\displaystyle \frac{d\tan x}{dx}}={\sec }^{2}x$
So simplifying the derivatives further with help of this,
$$\begin{gathered} {\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{1}{\tan x}}\times {\sec }^{2}x \\ \Rightarrow {\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{{\sec }^{2}x}{\tan x}} \end{gathered}$$
You can left it like this or should simplify it further in sine and cosine as follows
$$\begin{gathered} {\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{{\sec }^{2}x}{\tan x}} \\ \Rightarrow {\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{1}{{\displaystyle \frac{\sin x}{\cos x}}\times {\cos }^{2}x}} \\ \therefore {\displaystyle \frac{d\left(f(x)\right)}{dx}}={\displaystyle \frac{1}{\sin x\cos x}}=\csc x\sec x \end{gathered}$$
Therefore the desired derivative of $f(x)=\ln (\tan x)$ is equal to $\csc x\sec x$.

Note: When applying the chain rule in a more complex composite function then do the calculations or simplification stepwise, because chain rule also becomes complex for complex composite functions and your one mistake will change the whole answer.