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
$\begin{aligned} & a)\dfrac{{{x}^{2}}}{1-{{x}^{4}}} \\\ & b)\dfrac{2{{x}^{2}}}{1-{{x}^{4}}} \\\ & c)\dfrac{{{x}^{2}}}{2\left( 1-{{x}^{2}} \right)} \\\ & d)\text{ None of these}\text{. } \\\ \end{aligned}$

Answer 1

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,
$\begin{aligned} & y=\dfrac{1}{4}\left[ \log \left( 1+x \right)-\log \left( 1-x \right) \right]-\dfrac{1}{2}{{\tan }^{-1}}x \\\ & \Rightarrow y=\dfrac{1}{4}\log \left( 1+x \right)-\dfrac{1}{4}\log \left( 1-x \right)-\dfrac{1}{2}{{\tan }^{-1}}x \\\ \end{aligned}$
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,

& \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{aligned}$$ Now we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ $$\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{aligned} & \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{aligned}$$ Now again using the formula ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ we get, $$\begin{aligned} & \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{aligned}$$ 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 $\dfrac{d\left( \dfrac{f}{g} \right)}{dx}=\dfrac{f'g-g'f}{{{g}^{2}}}$ .