Question

Find the derivative of $\dfrac{d}{{dx}}\left( {\log \left( {\tan x} \right)} \right)$
$A)2\sec 2x$
$B)2\cos ec2x$
$C)\sec 2x$
$D)\cos ec2x$

Answer 1

Here we need to find a differentiation of $\log({\tan x})$ with respect to ‘x’. We can calculate the above differentiation with the help of the chain rule. Then we simplify the trigonometric function as our desires using the different trigonometric relations. While simplifying we will use trigonometric pythagorean identity that is ${\cos ^2}x + {\sin ^2}x = 1$.

Complete step by step answer:
The given function which we have to differentiate is
$h\left( x \right) = \log \left( {\tan x} \right)$
The above function is a composite function of the form of
$h\left( x \right) = f\left( {g\left( x \right)} \right)$
So to differentiate such a type of function we have to apply the chain rule. Which is shown below,
In calculus, the chain rule is a formula to compute the derivative of a composite function. That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite $fog$ the function which maps x to $f\left( {g\left( x \right)} \right)$ in terms of the derivatives of $f$ and $g$ and the product of functions as follows:
$\Rightarrow \left( {fog} \right)' = \left( {f'og} \right)g'$
Where alternatively, by letting
$\Rightarrow h = \left( {fog} \right)$
$\Rightarrow$ $h\left( x \right) = f\left( {g\left( x \right)} \right)$
One may express the chain rule in Lagrange's notation as
$\Rightarrow h'\left( x \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)$
Hence applying the chain rule to find the required answer ,
$\dfrac{d}{{dx}}\left( {\log \left( {\tan x} \right)} \right)$
$= \dfrac{1}{{\tan x}}\dfrac{d}{{dx}}\left( {\tan x} \right)$
We know that
$\tan x = \left( {\dfrac{{\sin x}}{{cox}}} \right)$
And, $\dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}}$
Hence we can write the derivative as shown below.
$= \dfrac{1}{{\tan x}}\dfrac{d}{{dx}}\left( {\dfrac{{\sin x}}{{\cos x}}} \right)$
$= \dfrac{1}{{\tan x}} \times \left( {\dfrac{{\cos x\dfrac{d}{{dx}}\left( {\sin x} \right) - \sin x\dfrac{d}{{dx}}\left( {\cos x} \right)}}{{{{\cos }^2}x}}} \right)$
$= \dfrac{{\cos x}}{{\sin x}} \times \dfrac{{\cos x.\cos x - \sin x\left( { - \sin x} \right)}}{{{{\cos }^2}x}}$
$= \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\sin x \times \cos x}}$
Now we know the relation between the trigonometric identify as
$\Rightarrow {\cos ^2}x + {\sin ^2}x = 1$
Therefore the above step can be written as,
$= \dfrac{1}{{\sin x \times \cos x}}$
Now multiplying ‘2’ in the denominator and divided by ‘2’ .We get,
$= \dfrac{2}{{2\sin x \times \cos x}}$
We know the trigonometric relation as,
$\Rightarrow \sin 2x = 2\sin x \times \cos x$
Therefore we get,
$= \dfrac{2}{{\sin 2x}}$
As we know that,
$\Rightarrow \dfrac{1}{{\sin 2x}} = \cos ec2x$
Therefore we get the final answer as,
$\dfrac{d}{{dx}}\left( {\log \left( {\tan x} \right)} \right) = 2\cos ec2x$
Hence, the correct option of the above differentiation is option (B) $2\cos ec2x$.

Note:
We have obtained the first derivative. If we differentiate it again we will have a second derivative, if we differentiate the second derivative we will have a third derivative and so on. If they asked us to find a second derivative we just need to differentiate the obtained solution again.