Question

If \[y=\left\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\}\] for – 1 < x < 1 then \[\dfrac{dy}{dx}\] is equal to

Answer 1

Hint: In order to solve this question, we should know about a few inverse trigonometric ratio derivatives like \[\dfrac{d}{dx}\left( {{\cot }^{-1}}x \right)=\dfrac{-1}{{{x}^{2}}+1}\]. We should also remember that the derivative of \[\sqrt{x}\], that is \[\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}},\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)\text{ and }\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]. By using these properties, we can solve this question.

Complete step-by-step answer:

In this question, we have been given an equation \[y=\left\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\}\] for – 1 < x < 1 and we have been asked to find \[\dfrac{dy}{dx}\]. To solve this question, we should have the knowledge of a few standard derivatives like \[\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}},\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)\text{ and }\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]. Now, to solve this, we will first find the derivative of \[{{\cot }^{-1}}\] and then the derivative of square root and then the derivative of \[\dfrac{1+x}{1-x}\]. So, we can write,
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\}\]
Now, we know that \[\dfrac{d}{dx}\left( {{\cot }^{-1}}x \right)=\dfrac{-1}{{{x}^{2}}+1}\]. So, for \[x=\sqrt{\dfrac{1+x}{1-x}}\], we get,
\[\dfrac{dy}{dx}=\dfrac{-1}{{{\left( \sqrt{\dfrac{1+x}{1-x}} \right)}^{2}}+1}.\dfrac{d}{dx}\left( \sqrt{\dfrac{1+x}{1-x}} \right)\]
Now, we know that \[{{\left( \sqrt{x} \right)}^{2}}=x\]. So, for \[x=\dfrac{1+x}{1-x}\], we get,
\[\dfrac{dy}{dx}=\dfrac{-1}{\dfrac{1+x}{1-x}+1}.\dfrac{d}{dx}\left( \sqrt{\dfrac{1+x}{1-x}} \right)\]
And we can further write it as
\[\dfrac{dy}{dx}=\dfrac{-1}{\dfrac{1+x+1-x}{1-x}}.\dfrac{d}{dx}\left( \sqrt{\dfrac{1+x}{1-x}} \right)\]
\[\dfrac{dy}{dx}=\dfrac{-1\left( 1-x \right)}{2}.\dfrac{d}{dx}\sqrt{\dfrac{1+x}{1-x}}\]
\[\dfrac{dy}{dx}=\left( \dfrac{x-1}{2} \right)\dfrac{d}{dx}\sqrt{\dfrac{1+x}{1-x}}\]
Now, we know that
\[\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}\]. So, for \[x=\dfrac{1+x}{1-x}\], we get,
\[\dfrac{dy}{dx}=\left( \dfrac{x-1}{2} \right).\dfrac{1}{2\sqrt{\dfrac{1+x}{1-x}}}\dfrac{d}{dx}\left( \dfrac{1+x}{1-x} \right)\]
And we can further write it as,
\[\dfrac{dy}{dx}=\left( \dfrac{x-1}{2} \right).\left( \dfrac{1}{2}\sqrt{\dfrac{1-x}{1+x}} \right)\dfrac{d}{dx}\left( \dfrac{1+x}{1-x} \right)\]
\[\dfrac{dy}{dx}=\dfrac{\left( x-1 \right)\sqrt{1-x}}{4\sqrt{1+x}}\dfrac{d}{dx}\left( \dfrac{1+x}{1-x} \right)\]
Now, we know that \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]. So, for u = 1 + x and v = 1 – x, we can write \[\dfrac{du}{dx}=1\text{ and }\dfrac{dv}{dx}=-1\]. Therefore, we get,
\[\dfrac{dy}{dx}=\dfrac{\left( x-1 \right)\sqrt{1-x}}{4\sqrt{1+x}}\left[ \dfrac{\left( 1-x \right)\left( 1 \right)-\left( 1+x \right)\left( -1 \right)}{{{\left( 1-x \right)}^{2}}} \right]\]
Now, we will simplify it further to get \[\dfrac{dy}{dx}\]. So, we get,
\[\dfrac{dy}{dx}=\dfrac{\left( x-1 \right)\sqrt{1-x}}{4\sqrt{1+x}}\left[ \dfrac{1-x+1+x}{{{\left( x-1 \right)}^{2}}} \right]\]
\[\dfrac{dy}{dx}=\dfrac{\sqrt{1-x}}{4\sqrt{1+x}}\left( \dfrac{2}{x-1} \right)\]
Now, we will write (x – 1) = – (1 – x). So, we will get,
\[\dfrac{dy}{dx}=\dfrac{-\sqrt{1-x}}{2\sqrt{1+x}}\left( \dfrac{1}{1-x} \right)\]
\[\dfrac{dy}{dx}=\dfrac{-{{\left( 1-x \right)}^{\dfrac{1}{2}-1}}}{2\sqrt{1+x}}\]
\[\dfrac{dy}{dx}=\dfrac{-{{\left( 1-x \right)}^{\dfrac{-1}{2}}}}{2\sqrt{1+x}}\]
And we know that \[{{x}^{\dfrac{-1}{2}}}=\dfrac{1}{\sqrt{x}}\]. So for x = 1 – x, we get,
\[\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{\left( 1+x \right)\left( 1-x \right)}}\]
And we can further write it as,
\[\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}\]
Hence, we can say that the derivative of \[y={{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}}\] for -1 < x < 1 is \[\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}\].

Note: While solving this question, we need to be very careful because there are chances of calculation mistakes. Also, we need to remember a few derivatives like \[\dfrac{d}{dx}\left( {{\cot }^{-1}}x \right)=\dfrac{-1}{{{x}^{2}}+1},\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}},\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)\text{ and }\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]. We might make a mistake while writing the derivatives of \[{{\cot }^{-1}}x\] and \[\dfrac{u}{v}\] type derivative. So, we have to be very focused and careful.