Question

d/dx cot⁻¹ √1+x²-1/x=......... (x ∈ R - {0}) ,Select Proper option from the given options.
(a) 1/1+x²
(b) 1/2(1+x²)
(c) 2/1+x²
(d) - 1/1+x²

Answer 1

Answer:

$\displaystyle\,\frac{dy}{dx}=\frac{-1}{2(1+x^2)}$

Step-by-step explanation:

$\displaystyle\text{Let }y=cot^{-1}(\frac{\sqrt{1+x^2}-1}{x})$

$\text{put }x=tan\theta\,\implies\:\theta=tan^{-1}x$

$\displaystyle\,y=cot^{-1}(\frac{\sqrt{1+tan^2\theta}-1}{tan\theta})$

$\displaystyle\,y=cot^{-1}(\frac{\sqrt{sec^2\theta}-1}{tan\theta})$

$\displaystyle\,y=cot^{-1}(\frac{sec\theta-1}{tan\theta})$

$\displaystyle\,y=cot^{-1}(\frac{\frac{1}{cos\theta}-1}{\frac{sin\theta}{cos\theta}})$

$\displaystyle\,y=cot^{-1}(\frac{\frac{1-cos\theta}{cos\theta}}{\frac{sin\theta}{cos\theta}})$

$\displaystyle\,y=cot^{-1}(\frac{1-cos\theta}{sin\theta})$

Using

$\boxed{\begin{minipage}{5cm} \bf\,cosA=1-2sin^2\frac{A}{2}\\ \\ \implies1-cosA=2\,sin^2\frac{A}{2}\\ \\sinA=2\,sin\frac{A}{2}\,cos\frac{A}{2} \end{minipage}}$

$\displaystyle\,y=cot^{-1}(\frac{2\,sin^2\frac{\theta}{2}}{2\,sin\frac{\theta}{2}\,cos\frac{\theta}{2}})$

$\displaystyle\,y=cot^{-1}(\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}})$

$\displaystyle\,y=cot^{-1}(tan\frac{\theta}{2})$

$\displaystyle\,y=cot^{-1}(cot(\frac{\pi}{2}-\frac{\theta}{2}))$

$\displaystyle\,y=\frac{\pi}{2}-\frac{1}{2}\theta$

$\displaystyle\,y=\frac{\pi}{2}-\frac{1}{2}tan^{-1}x$

$\text{Differentiate with respect to x}$

$\displaystyle\,\frac{dy}{dx}=-\frac{1}{2}(\frac{1}{1+x^2})$

$\displaystyle\,\implies\boxed{\frac{dy}{dx}=\frac{-1}{2(1+x^2)}}$

Answer 2

Answer:

$\frac{-1}{2(1+x^{2} )}$

Step-by-step explanation:

Let us assume that  y= Cot¬1($\frac{\sqrt{1+x^{2} }-1}{x}$).

Differentiating both sides with respect to x, we obtain

$\frac{dy}{dx}= -\frac{1}{1+(\frac{\sqrt{1+x^{2}}-1}{x})^{2}}(\frac{d}{dx}(\frac{\sqrt{1+x^{2}}-1}{x}))$

{ Since, we have the formula d/dx(Cot¬1 x)=$-\frac{1}{1+x^{2} }$}

= $-\frac{1}{1+(\frac{x^{2}+2-2\sqrt{1+x^{2}}}{x^{2} })}\frac{x(\frac{2x}{2\sqrt{1+x^{2} } } )-(\sqrt{1+x^{2} }-1 )}{x^{2} }$

= $-\frac{x^{2}}{2x^{2}+2-2\sqrt{1+x^{2}}}\frac{\frac{x^{2} }{\sqrt{1+x^{2} } }-(\sqrt{1+x^{2} }-1 ) }{x^{2} }$

= $-\frac{x^{2}}{2(x^{2}+1-\sqrt{1+x^{2}})} .\frac{x^{2}-1-x^{2}+\sqrt{1+x^{2}}}{x^{2}\sqrt{1+x^{2}}}$

= $-\frac{1}{2\sqrt{1+x^{2} }(\sqrt{1+x^{2}}-1)}. \frac{(\sqrt{1+x^{2} }-1 )}{\sqrt{1+x^{2} } }$

= $\frac{-1}{2(1+x^{2} )}$

(Answer)