
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
483k+ views
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.
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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Basicity of sulphurous acid and sulphuric acid are

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE
