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Question: If \[y=\left\\{ {{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} \right\\}\] for – 1 < x < 1 then \[\dfrac{dy}{...

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

Explanation

Solution

Hint: In order to solve this question, we should know about a few inverse trigonometric ratio derivatives like ddx(cot1x)=1x2+1\dfrac{d}{dx}\left( {{\cot }^{-1}}x \right)=\dfrac{-1}{{{x}^{2}}+1}. We should also remember that the derivative of x\sqrt{x}, that is ddxx=12x,ddx(f(g(x)))=f(g(x)).g(x) and ddx(uv)=vdudxudvdxv2\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 dydx\dfrac{dy}{dx}. To solve this question, we should have the knowledge of a few standard derivatives like ddxx=12x,ddx(f(g(x)))=f(g(x)).g(x) and ddx(uv)=vdudxudvdxv2\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 cot1{{\cot }^{-1}} and then the derivative of square root and then the derivative of 1+x1x\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 ddx(cot1x)=1x2+1\dfrac{d}{dx}\left( {{\cot }^{-1}}x \right)=\dfrac{-1}{{{x}^{2}}+1}. So, for x=1+x1xx=\sqrt{\dfrac{1+x}{1-x}}, we get,
dydx=1(1+x1x)2+1.ddx(1+x1x)\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 (x)2=x{{\left( \sqrt{x} \right)}^{2}}=x. So, for x=1+x1xx=\dfrac{1+x}{1-x}, we get,
dydx=11+x1x+1.ddx(1+x1x)\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
dydx=11+x+1x1x.ddx(1+x1x)\dfrac{dy}{dx}=\dfrac{-1}{\dfrac{1+x+1-x}{1-x}}.\dfrac{d}{dx}\left( \sqrt{\dfrac{1+x}{1-x}} \right)
dydx=1(1x)2.ddx1+x1x\dfrac{dy}{dx}=\dfrac{-1\left( 1-x \right)}{2}.\dfrac{d}{dx}\sqrt{\dfrac{1+x}{1-x}}
dydx=(x12)ddx1+x1x\dfrac{dy}{dx}=\left( \dfrac{x-1}{2} \right)\dfrac{d}{dx}\sqrt{\dfrac{1+x}{1-x}}
Now, we know that
ddxx=12x\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}. So, for x=1+x1xx=\dfrac{1+x}{1-x}, we get,
dydx=(x12).121+x1xddx(1+x1x)\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,
dydx=(x12).(121x1+x)ddx(1+x1x)\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)
dydx=(x1)1x41+xddx(1+x1x)\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 ddx(uv)=vdudxudvdxv2\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 dudx=1 and dvdx=1\dfrac{du}{dx}=1\text{ and }\dfrac{dv}{dx}=-1. Therefore, we get,
dydx=(x1)1x41+x[(1x)(1)(1+x)(1)(1x)2]\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 dydx\dfrac{dy}{dx}. So, we get,
dydx=(x1)1x41+x[1x+1+x(x1)2]\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]
dydx=1x41+x(2x1)\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,
dydx=1x21+x(11x)\dfrac{dy}{dx}=\dfrac{-\sqrt{1-x}}{2\sqrt{1+x}}\left( \dfrac{1}{1-x} \right)
dydx=(1x)12121+x\dfrac{dy}{dx}=\dfrac{-{{\left( 1-x \right)}^{\dfrac{1}{2}-1}}}{2\sqrt{1+x}}
dydx=(1x)1221+x\dfrac{dy}{dx}=\dfrac{-{{\left( 1-x \right)}^{\dfrac{-1}{2}}}}{2\sqrt{1+x}}
And we know that x12=1x{{x}^{\dfrac{-1}{2}}}=\dfrac{1}{\sqrt{x}}. So for x = 1 – x, we get,
dydx=12(1+x)(1x)\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{\left( 1+x \right)\left( 1-x \right)}}
And we can further write it as,
dydx=121x2\dfrac{dy}{dx}=\dfrac{-1}{2\sqrt{1-{{x}^{2}}}}
Hence, we can say that the derivative of y=cot11+x1xy={{\cot }^{-1}}\sqrt{\dfrac{1+x}{1-x}} for -1 < x < 1 is dydx=121x2\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 ddx(cot1x)=1x2+1,ddxx=12x,ddx(f(g(x)))=f(g(x)).g(x) and ddx(uv)=vdudxudvdxv2\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 cot1x{{\cot }^{-1}}x and uv\dfrac{u}{v} type derivative. So, we have to be very focused and careful.