Question

If \[y=\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}-\dfrac{1}{2}{{\tan }^{-1}}x\] then \[\dfrac{dy}{dx}\] is
$\begin{align}
  & a)\dfrac{{{x}^{2}}}{1-{{x}^{4}}} \\
 & b)\dfrac{2{{x}^{2}}}{1-{{x}^{4}}} \\
 & c)\dfrac{{{x}^{2}}}{2\left( 1-{{x}^{2}} \right)} \\
 & d)\text{ None of these}\text{. } \\
\end{align}$

Answer 1

Hint: Now we are given with an equation. Let us first consider $f\left( x \right)=\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}$ using the properties of log which says $\log {{a}^{n}}=n\log a$ and $\log \dfrac{a}{b}=\log a-\log b$ we simplify the equation. Substitute this value of $\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}$ in the given equation and differentiate the whole equation. Now we know that the differentiation of $\log \left( a+bx \right)=\dfrac{1}{b}\log \left( \dfrac{1}{a+bx} \right)$ and the differentiation of ${{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$ . Now we will simplify the equation obtained and hence arrive at the required answer.

Complete step-by-step answer:
Now we are given that \[y=\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}-\dfrac{1}{2}{{\tan }^{-1}}x\]
Let us let us say $f\left( x \right)=\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}$ .
Now we know that $\log {{a}^{n}}=n\log a$ .
Hence using this property we can write \[\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}=\dfrac{1}{4}\left( \log \left( \dfrac{1+x}{1-x} \right) \right).........................\left( 1 \right)\]
Now \[\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}=\dfrac{1}{4}\left( \log \left( \dfrac{1+x}{1-x} \right) \right)\]
Now we also know that $\log \dfrac{a}{b}=\log a-\log b$ .
Hence using this property in the obtained equation we get, \[\dfrac{1}{4}\left( \log \left( \dfrac{1+x}{1-x} \right) \right)=\dfrac{1}{4}\left[ \log \left( 1+x \right)-\log \left( 1-x \right) \right]\] .
Now let us substitute the value from above equation in equation (1) we get,
$\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}=\dfrac{1}{4}\left[ \log \left( 1+x \right)-\log \left( 1-x \right) \right]................\left( 2 \right)$
Now let us again consider the given equation \[y=\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}-\dfrac{1}{2}{{\tan }^{-1}}x\]
Substituting the value of $\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}$ from equation (2) in the above equation we get,
$\begin{align}
  & y=\dfrac{1}{4}\left[ \log \left( 1+x \right)-\log \left( 1-x \right) \right]-\dfrac{1}{2}{{\tan }^{-1}}x \\
 & \Rightarrow y=\dfrac{1}{4}\log \left( 1+x \right)-\dfrac{1}{4}\log \left( 1-x \right)-\dfrac{1}{2}{{\tan }^{-1}}x \\
\end{align}$
Now differentiating on both sides we get,
$\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{4}\log \left( 1+x \right) \right)}{dx}-\dfrac{d\left( \dfrac{1}{4}\log \left( 1-x \right) \right)}{dx}-\dfrac{d\left( \dfrac{1}{2}{{\tan }^{-1}}x \right)}{dx}$
Now we know that $\dfrac{d\left( cf\left( x \right) \right)}{dx}=c\dfrac{d\left( f\left( x \right) \right)}{dx}$ where c is the constant. Hence using this property we get,
$\dfrac{dy}{dx}=\dfrac{1}{4}\dfrac{d\left( \log \left( 1+x \right) \right)}{dx}-\dfrac{1}{4}\dfrac{d\left( \log \left( 1-x \right) \right)}{dx}-\dfrac{1}{2}\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}$ .
Now we know that differentiation of $\log \left( a+bx \right)=\dfrac{1}{b}\log \left( \dfrac{1}{a+bx} \right)$ Hence we get,
\[\dfrac{dy}{dx}=\dfrac{1}{4}\times \dfrac{1}{1+x}-\dfrac{1}{4}\times \dfrac{1}{1-x}\times \left( -1 \right)-\dfrac{1}{2}\times \dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}\] .
Now we also know that \[\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}\] .
Hence we get,
\[\dfrac{dy}{dx}=\dfrac{1}{4}\times \left( \dfrac{1}{1+x}+\dfrac{1}{1-x} \right)-\dfrac{1}{2}\times \dfrac{1}{1+{{x}^{2}}}\]
Now taking LCM of the given fractions we get,
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{4}\times \left( \dfrac{1-x+1+x}{\left( 1+x \right)\left( 1-x \right)} \right)-\dfrac{1}{2\left( 1+{{x}^{2}} \right)} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{4}\times \left( \dfrac{2}{\left( 1+x \right)\left( 1-x \right)} \right)-\dfrac{1}{2\left( 1+{{x}^{2}} \right)} \\
\end{align}\]
Now we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\times \left( \dfrac{1}{1-{{x}^{2}}} \right)-\dfrac{1}{2\left( 1+{{x}^{2}} \right)}\]
Now taking LCM we get,
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{2\left( 1+{{x}^{2}} \right)-2\left( 1-{{x}^{2}} \right)}{2\left( 1-{{x}^{2}} \right)2\left( 1+{{x}^{2}} \right)} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{2\left( 1+{{x}^{2}}-1+{{x}^{2}} \right)}{4\left( 1-{{x}^{2}} \right)\left( 1+{{x}^{2}} \right)} \right) \\
\end{align}\]
Now again using the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ we get,
\[\begin{align}
  & \dfrac{dy}{dx}=\left( \dfrac{2{{x}^{2}}}{2\left( 1-{{x}^{4}} \right)} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{1-{{x}^{4}}} \\
\end{align}\]
Hence we have the value of $\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{1-{{x}^{2}}}$

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

Note: Now we also know the chain rule of differentiation which says $\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( f\left( x \right) \right).g'\left( x \right)$ . We can also use this to solve the given equation. We can use chain rule to differentiate the function $f\left( x \right)=\log {{\left( \dfrac{1+x}{1-x} \right)}^{\dfrac{1}{4}}}$ instead of simplifying with the help of properties of log. Also note that to do so we will require the formula $\dfrac{d\left( \dfrac{f}{g} \right)}{dx}=\dfrac{f'g-g'f}{{{g}^{2}}}$ .