Question

If y = $\sqrt{x^2+1}- \log \bigg \lgroup \frac{1}{x} + \sqrt{1+\frac{1}{x^2}} \bigg \rgroup$ ,then find dy/dx

Answer 1

We have,

$\sf{y = \sqrt{x^2+1}- \log \bigg \lgroup \frac{1}{x} + \sqrt{1+\frac{1}{x^2}} \bigg \rgroup}$

$\sf{ \implies y = \sqrt{x^2+1}- \log \bigg \lgroup \frac{1 + \sqrt{ {x}^{2} + 1} }{x} \bigg \rgroup}$

$\sf{ \implies y = \sqrt{x^2+1}- \log \bigg \lgroup 1 + \sqrt{ {x}^{2} + 1} \bigg \rgroup} + \log x$

On differentiating w.r.t. x, we have

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

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

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

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

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

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

$\sf = \frac{ {x}^{2} + 1}{x \sqrt{ {x}^{2} + 1 } }$

$\sf= \frac{ \sqrt{ {x}^{2} + 1} }{x}$

Answer 2

Answer:

$\frac{\text{d}x}{\text{d}y} = \frac{\sqrt{x^2+1} }{x}$

Step-by-step explanation:

$\bold{We\:have,}=y = \sqrt{x^2+1}- \log \bigg \lgroup \frac{1}{x} + \sqrt{1+\frac{1}{x^2}} \bigg \rgroup$

$\Rightarrow y = \sqrt{x^2+1}- \log \bigg \lgroup \frac{1+\sqrt{x^2+1} }{x} \bigg \rgroup$

$\Rightarrow y = \sqrt{x^2+1}- \log \bigg \lgroup 1+\sqrt{x^2+1}+\log x \bigg \rgroup$

On differentiating w.r.t. x, we have,

$\frac{\text{d}x}{\text{d}y} = \frac{\sqrt{x^2+1} }{x}$

$\red{:\longmapsto}\green{\boxed{\frac{dy}{dx} = \frac{\sqrt{x^2+1} }{x}}$