If y-log x-x 2-12- show that 1-x 2 d 2 y d x 2-x d y d x-2

Here, y=logx+x2+12Differentiating #160;w.r.t. #160;x, #160;we #160;getdydx=2logx+x2+1x+x2+1 #215;1+2x2x2+1 #8658;dydx=2logx+x2+1x+x2+1 #215;x2+1+x ...

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Byju's Answer

Standard XII

Mathematics

Log Function

If y=log x+x ...

Question

If $y = {[\log (x + \sqrt{x^{2} + 1})]}^{2}$ , show that $(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2$ .

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Solution

Here,
$y = {[\log (x + \sqrt{x^{2} + 1})]}^{2} Differentiating w . r . t . x, we get \frac{d y}{d x} = \frac{2 \log (x + \sqrt{x^{2} + 1})}{(x + \sqrt{x^{2} + 1})} \times (1 + \frac{2 x}{2 \sqrt{x^{2} + 1}}) \Rightarrow \frac{d y}{d x} = \frac{2 \log (x + \sqrt{x^{2} + 1})}{(x + \sqrt{x^{2} + 1})} \times (\frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1}}) \Rightarrow \frac{d y}{d x} = \frac{2 \log (x + \sqrt{x^{2} + 1})}{\sqrt{x^{2} + 1}} Differentiating again w . r . t . x, we get \frac{d^{2} y}{d x^{2}} = \frac{2 - \frac{2 x \log (x + \sqrt{x^{2} + 1})}{\sqrt{x^{2} + 1}}}{x^{2} + 1} \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{2 - x \frac{d y}{d x}}{x^{2} + 1} \Rightarrow (x^{2} + 1) \frac{d^{2} y}{d x^{2}} = 2 - x \frac{d y}{d x} \Rightarrow (x^{2} + 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2$

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If y=log x+x2+12, show that 1+x2d2ydx2+xdydx=2.

Here, y=logx+x2+12Differentiating w.r.t. x, we getdydx=2logx+x2+1x+x2+1×1+2x2x2+1⇒dydx=2logx+x2+1x+x2+1×x2+1+xx2+1⇒dydx=2logx+x2+1x2+1Differentiating again w.r.t. x, we getd2ydx2=2-2xlogx+x2+1x2+1x2+1⇒d2ydx2=2-xdydxx2+1⇒x2+1d2ydx2=2-xdydx⇒x2+1d2ydx2+xdydx=2

If $y = {[\log (x + \sqrt{x^{2} + 1})]}^{2}$ , show that $(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = 2$ .