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.