Question

Differential coefficient of $\log \left( {\sin x} \right)$ is:
$
  A.{\text{ }}\cos x \\
  B.{\text{ }}\tan x \\
  C.{\text{ cosec }}x \\
  D.{\text{ }}\cot x \\
$

Answer 1

Hint: In this question use the property of log differentiation which is differentiation of $\log \left( {ax} \right)$ w.r.t. x is $\dfrac{1}{{ax}}\left( {\dfrac{d}{{dx}}ax} \right)$, so use this concept to reach the solution of the question.

Complete step-by-step answer:
Given equation is

$\log \left( {\sin x} \right)$

Let, $y = \log \left( {\sin x} \right)$

Now we have to find out the differentiation of the above equation.

So differentiate it w.r.t. x we have,

$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\log \left( {\sin x} \right)} \right)$

Now we all know that the differentiation of $\log \left( {ax} \right)$ w.r.t. x is

$\dfrac{1}{{ax}}\left( {\dfrac{d}{{dx}}ax} \right)$, so use this property of log differentiation in

above equation we have,

$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin x}}\left( {\dfrac{d}{{dx}}\sin x} \right)$

Now again re-differentiate sin x w.r.t. x we have, (differentiation of sin x is cos x)

 $\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin x}}\left( {\cos x} \right)$

Now as we know $\dfrac{{\cos x}}{{\sin x}} = \cot x$ so, substitute this value in above equation we have,
$\dfrac{{dy}}{{dx}} = \cot x$

So, this is the required differentiation of $\log \left( {\sin x} \right)$.

Hence option (D) is correct.

Note: In such types of questions the key concept we have to remember is that always recall how to differentiate log which is stated above then using this property differentiate it then re-differentiate sin x and simplify or we can calculate the differentiation of $\log \left( {\sin x} \right)$ by using first principle method which is given as $\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}$.