Question

If $ y = \ln \sqrt {\tan x} $ then the value of $ \dfrac{{dy}}{{dx}} $ ​ at $ x = \dfrac{\pi }{4} $ ​ is-
A. $ \infty $
B. 1
C. 0
D. $ \dfrac{1}{2} $

Answer 1

Hint: Here the function that is given to us is an impure function consisting of trigonometric function, the under root function and logarithmic function. To find the derivative with respect to ‘x’ we have to use the chain rule. We will be using the basic differentiation formulas here as listed below.
a) If a function ‘f’ is a mixture of various function say $ f(x) = {g_1}({g_2}({g_3}(...{g_n}(x)...))) $ then the chain rule says that the differentiation of ‘f’ with respect to ‘x’ is \[{f^1}(x) = \dfrac{{df(x)}}{{dx}} = \dfrac{{d\left[ {{g_1}({g_2}({g_3}(...{g_n}(x)...)))} \right]}}{{d\left[ {{g_2}({g_3}(...{g_n}(x)...))} \right]}} \times \dfrac{{d\left[ {{g_2}({g_3}(...{g_n}(x)...))} \right]}}{{d\left[ {{g_3}(...{g_n}(x)...)} \right]}} \times \dfrac{{d\left[ {{g_3}(...{g_n}(x)...)} \right]}}{{d\left[ {(...{g_n}(x)...} \right]}} \times ... \times \dfrac{{d\left[ {{g_n}(x)} \right]}}{{dx}}\]
b) The basic differentiation formulas are,
 $ \dfrac{{d\tan x}}{{dx}} = {\sec ^2}x $
 $ \dfrac{{d\sqrt x }}{{dx}} = \dfrac{1}{{2\sqrt x }} $
 $ \dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x} $
c) From trigonometry we have,
 $ \dfrac{1}{{\tan x}} = \dfrac{{\cos x}}{{\sin x}} $
 $ \sec x = \dfrac{1}{{\cos x}} $
\[2\sin x\cos x = \sin 2x\]
\[\csc \left( {\dfrac{\pi }{2}} \right) = 1\]

Complete step-by-step answer:
Here the function is given as $ y = \ln \sqrt {\tan x} $
So, we have the formula $ \dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x} $
As here the function is $ \ln \sqrt {\tan x} $ in place of ‘x’ we have to use the chain rule.
So we will write the differentiation as
 $ y = \ln \sqrt {\tan x} $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\ln \sqrt {\tan x} } \right) $
Applying chain rule:
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{d\sqrt {\tan x} }}\left( {\ln \sqrt {\tan x} } \right) \times \dfrac{{d\sqrt {\tan x} }}{{dx}} $
As we have, $ \dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x} $ so we can write, $ \dfrac{{d\ln \sqrt {\tan x} }}{{d\sqrt {\tan x} }} = \dfrac{1}{{\sqrt {\tan x} }} $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\tan x} }} \times \dfrac{{d\sqrt {\tan x} }}{{dx}} $
Applying chain rule:
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\tan x} }} \times \dfrac{{d\sqrt {\tan x} }}{{d\tan x}} \times \dfrac{{d\tan x}}{{dx}} $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\tan x} }} \times \dfrac{1}{{2\sqrt {\tan x} }} \times \dfrac{{d\tan x}}{{dx}} $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\tan x} }} \times \dfrac{1}{{2\sqrt {\tan x} }} \times {\sec ^2}x $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\tan x}} \times {\sec ^2}x $
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{2\sin x}} \times \dfrac{1}{{{{\cos }^2}x}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sin x}} \times \dfrac{1}{{\cos x}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sin x\cos x}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin 2x}} = \csc 2x\]
Now we are asked to find the value when $ x = \dfrac{\pi }{4} $
Now we have,
\[{\left[ {\dfrac{{dy}}{{dx}}} \right]_{\dfrac{\pi }{4}}} = \csc \left( {2 \times \dfrac{\pi }{4}} \right) = \csc \dfrac{\pi }{2} = 1\]
Hence the final answer is 1.
So, the correct answer is “1”.

Note: The chain rule is a simple yet long process to compute the differentiation of impure functions. Whenever you see a function not in pure form always use this formula for chain rule stated in the formula section. Make sure you do it step by step as small mistakes can ruin the whole differentiation process.