Lf y-logx-1-x2- then y-0 is

The correct option is C 0 y=log(x+√(1+x^2)) y'=1+x√(1+x^2)x+√(1+x^2)=1√(1+x^2) y”= 12(2x(1+x^2)^3/2) y”(0)=0 ...

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Fundamental Laws of Logarithms

lf y=logx+√...

Question

lf $y = log (x + \sqrt{1 + x^{2}})$ , then $y^{^{''}} (0)$ is

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Similar questions

y = \frac{log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}

(1 + x^{2}) y_{1} + x y =

x^{2} y_{2} + x y_{1} + y = 0

y = cos (m {sin}^{- 1} x)

(1 - x^{2}) y_{2} - x y_{1} + m^{2} y = 0

log y = {tan}^{- 1} x

(1 + x^{2}) \frac{d^{2} y}{d x^{2}} + (2 x - 1) \frac{d y}{d x} = 0

y = \frac{1}{2}

(\frac{1}{2_{X} \sqrt{1 + x^{2}}})

x =

y = x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots . \infty

y - \frac{1}{2} y^{2} + \frac{1}{3} y^{3} - \frac{1}{4} y^{4} + \dots . \infty =

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