If ![y = (x+\sqrt{1+x^2})^n, then (1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx} is y = (x+\sqrt{1+x^2})^n, then (1+x^2)\frac{d^2y}{dx^2}+x\frac{dy}{dx} is](https://tex.z-dn.net/?f=y+%3D+%28x%2B%5Csqrt%7B1%2Bx%5E2%7D%29%5En%2C+then+%281%2Bx%5E2%29%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%2Bx%5Cfrac%7Bdy%7D%7Bdx%7D+is)
(a) n²y
(b) - n²y
(c) -y
(d) 2x²y
Anonymous:
-n²y
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