Question

Differential co – efficient of \[\log \left( \sin x \right)\] with respect to x is: -
(a) \[\cot x\]
(b) \[\csc x\]
(c) \[\tan x\]
(d) \[\sec x\]

Answer 1

Hint: Understand the meaning of the term differential co – efficient. Use the chain rule of differentiation to find the derivative of the given function of \[\log \left( \sin x \right)\]. First differentiate \[\log \left( \sin x \right)\] with respect to \[\sin x\] and then differentiate \[\sin x\] with respect to x and take their product to get the answer.

Complete step-by-step answer:
Here, we have been provided with a function \[\log \left( \sin x \right)\] and we have to find its differential co – efficient with respect to x. But first we need to know the meaning of the term ‘differential co – efficient’.
In mathematics and physics, the differential co – efficient of a function g (x) is the term used for the derivative of g (x). So, we have,
\[\Rightarrow \]Differential co – efficient of g (x) = \[\dfrac{d\left( g\left( x \right) \right)}{dx}\].
Now, let us come to the question. Here, the function is \[\log \left( \sin x \right)\], and we have to find its derivative with respect to x.
The function \[\log \left( \sin x \right)\] can be written in the form \[f\left( g\left( x \right) \right)\] which is called a composite function. So, we have,
\[\begin{align}
  & \Rightarrow f\left( x \right)=\log x \\
 & \Rightarrow g\left( x \right)=\sin x \\
 & \Rightarrow f\left( g\left( x \right) \right)=\log \left( \sin x \right) \\
\end{align}\]
Now, we know that the derivative of a composite function \[f\left( g\left( x \right) \right)\] can be given by using chain rule of differentiation.
\[\Rightarrow \dfrac{df\left( g\left( x \right) \right)}{dx}=\dfrac{df\left( g\left( x \right) \right)}{d\left( g\left( x \right) \right)}\times \dfrac{d\left( g\left( x \right) \right)}{dx}\]
So, using the above chain rule formula to differentiate \[\log \left( \sin x \right)\] we get,
\[\Rightarrow \dfrac{d\left[ \log \left( \sin x \right) \right]}{dx}=\dfrac{d\left[ \log \left( \sin x \right) \right]}{d\left[ \sin x \right]}\times \dfrac{d\left[ \sin x \right]}{dx}\] - (1)
Now, we know that the derivative of \[\log x\] is \[\dfrac{1}{x}\] and \[\sin x\] is \[\cos x\], so we have,
\[\Rightarrow \dfrac{d\left[ \log \left( \sin x \right) \right]}{d\left[ \sin x \right]}=\dfrac{1}{\sin x}\]
And, \[\dfrac{d\left[ \sin x \right]}{d\left( x \right)}=\cos x\]
Therefore, substituting these values in equation (1), we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ \log \left( \sin x \right) \right]}{dx}=\dfrac{1}{\sin x}\times \cos x \\
 & \Rightarrow \dfrac{d\left[ \log \left( \sin x \right) \right]}{dx}=\dfrac{\cos x}{\sin x} \\
\end{align}\]
Applying the conversion, \[\dfrac{\cos x}{\sin x}=\cot x\], we get,
\[\Rightarrow \dfrac{d\left[ \log \left( \sin x \right) \right]}{dx}=\cot x\]

So, the correct answer is “Option (a)”.

Note: One must know the meaning of the term differential co – efficient to solve the question. This term was used for differentiation however in the process of integration also there are certain other terms used for integral, which are anti derivative and primitive. These are old terms and outdated however sometimes used to confuse us. You must remember the chain rule of differentiation used for the composite functions.