Question

Differentiate \[\sqrt {\tan x} \] with respect to $x$.
a) \[\dfrac{{{{\sec }^2}x}}{{2\sqrt {\tan x} }}\]
b) \[\dfrac{{ - {{\sec }^2}x}}{{2\sqrt {\tan x} }}\]
c) \[\dfrac{{{{\csc }^2}x}}{{2\sqrt {\tan x} }}\]
d) \[\dfrac{{ - {{\csc }^2}x}}{{2\sqrt {\tan x} }}\]

Answer 1

Hint: We need to differentiate \[\sqrt {\tan x} \], which can be written as \[{(\tan x)^{\dfrac{1}{2}}}\]. We will use the chain rule to differentiate the given function. Chain rule is applied on the composite functions. We will consider two functions \[f\] and \[g\]. Then, we will write the given function as a composition of these two functions. Chain Rule for differentiation of composite function can be defined as:
\[\dfrac{d}{{dx}}\left( {fog(x)} \right) = f'(g(x)) \times g'(x)\] , where \[\dfrac{d}{{dx}}(g(x)) = g'(x)\].

Complete step by step solution:
We need to differentiate \[\sqrt {\tan x} \].
\[\sqrt {\tan x} \] can be written as \[{(\tan x)^{\dfrac{1}{2}}}\]
Considering, \[f(x) = {x^{\dfrac{1}{2}}}\] and \[g(x) = \tan x\], we get
\[fog(x) = f(g(x)) = {(g(x))^{\dfrac{1}{2}}} = {(\tan x)^{\dfrac{1}{2}}}\] as \[g(x) = \tan x\]
Using Chain Rule in the for the above function, we get
\[\dfrac{d}{{dx}}\left( {{{(\tan x)}^{\dfrac{1}{2}}}} \right) = \dfrac{d}{{dx}}\left( {fog(x)} \right) = f'(g(x)) \times g'(x)\], where\[\dfrac{d}{{dx}}(g(x)) = g'(x)\] ----(1)
Now, we have
\[f'(x) = \dfrac{d}{{dx}}\left( {f(x)} \right) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)\]
As we know, \[\left( {\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}} \right)\]. So,
\[f'(x) = \dfrac{1}{2}\left( {{x^{\dfrac{1}{2} - 1}}} \right)\]
Taking LCM of the powers of \[x\].
\[f'(x) = \dfrac{1}{2}\left( {{x^{\dfrac{{1 - 2}}{2}}}} \right)\]
\[f'(x) = \dfrac{1}{2}\left( {{x^{ - \dfrac{1}{2}}}} \right)\]
As we have to find \[f'(g(x))\], we will replace \[x\] by \[g(x)\] in the above equation. Hence,
\[f'(g(x)) = \dfrac{1}{2}\left( {{{(g(x))}^{ - \dfrac{1}{2}}}} \right)\]
Now, Substituting the value of i.e. \[g(x) = \tan x\]
\[f'(g(x)) = \dfrac{1}{2}\left( {{{(\tan x)}^{ - \dfrac{1}{2}}}} \right)\]
Using the properties of exponential power, we know \[\left( {{x^{ - n}} = \dfrac{1}{{{x^n}}}} \right)\]. Hence the above equation becomes
\[f'(g(x)) = \dfrac{1}{2}\left( {\dfrac{1}{{{{(\tan x)}^{\dfrac{1}{2}}}}}} \right)\]
We know, \[\left( {{x^{\dfrac{1}{2}}} = \sqrt x } \right)\]. Hence, using this, we get
\[f'(g(x)) = \dfrac{1}{2}\left( {\dfrac{1}{{\sqrt {\tan x} }}} \right)\] -----(2)
Differentiating \[g(x)\] with respect to \[x\]
\[\dfrac{d}{{dx}}(g(x)) = g'(x) = \dfrac{d}{{dx}}\left( {\tan x} \right)\]
We know the differentiation formula for \[\tan x\] i.e. \[\left( {\dfrac{d}{{dx}}\left( {\tan x} \right) = {{\sec }^2}x} \right)\]
\[g'(x) = {\sec ^2}x\] ---(3)
Using (2) and (3) in (1), we get
\[\dfrac{d}{{dx}}\left( {\sqrt {\tan x} } \right) = \dfrac{d}{{dx}}\left( {{{(\tan x)}^{\dfrac{1}{2}}}} \right) = \dfrac{1}{2}\left( {\dfrac{1}{{\sqrt {\tan x} }}} \right) \times {\sec ^2}x\]
\[\dfrac{d}{{dx}}\left( {\sqrt {\tan x} } \right) = \dfrac{1}{2} \times \dfrac{1}{{\sqrt {\tan x} }} \times {\sec ^2}x\]
\[\dfrac{d}{{dx}}\left( {\sqrt {\tan x} } \right) = \dfrac{{{{\sec }^2}x}}{{2\sqrt {\tan x} }}\]
\[\therefore \]Differentiating \[\sqrt {\tan x} \] with respect to \[x\] we get \[\dfrac{{{{\sec }^2}x}}{{2\sqrt {\tan x} }}\].
\[\therefore \] The correct option is (a).

Note: We need to be very careful when we are applying the chain rule and when we decide two functions for a composite function. Composition of functions needs to be done with full presence of mind. Also, we need to remember the formulas for differentiation of functions thoroughly. While we are applying chain rule, we need to be very careful with the first term of the right hand side i.e. \[f'(g(x))\]. We have to differentiate the function \[f(x)\] with respect to \[x\] and then replace \[x\] by \[g(x)\] after obtaining the value of \[f'(x)\]. We don’t have to Differentiate \[g(x)\] in this step.