If y=axn+1+bxn+1,thenx2d2ydx2 is equal to

y=axn+1+bxn+1dydx=an+1xn+bn+1xn=n+1axn+bxnd2ydx2=n+1naxn-1+bnxn-1=nn+1axn-1+bxn-1x2d2ydx2=nn+1axn+1+bxn+1=nn+1y

If y=axn+1+bxn+1,thenx2d2ydx2 is equal to

y=axn+1+bxn+1dydx=an+1xn+bn+1xn=n+1axn+bxnd2ydx2=n+1naxn-1+bnxn-1=nn+1axn-1+bxn-1x2d2ydx2=nn+1axn+1+bxn+1=nn+1y

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If $y = a x^{n + 1} + b x^{n + 1}, then x^{2} \frac{d^{2} y}{d x^{2}}$ is equal to

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$n (n - 1) y$

$n (n + 1) y$

$n y$

$n^{2} y$

answer is B.

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Detailed Solution

$y = a x^{n + 1} + b x^{n + 1}$

$\frac{d y}{d x} = a (n + 1) x^{n} + b (n + 1) x^{n} = (n + 1) (a x^{n} + b x^{n})$

$\frac{d^{2} y}{d x^{2}} = (n + 1) (n a x^{n - 1} + b n x^{n - 1}) = (n) (n + 1) (a x^{n - 1} + b x^{n - 1})$

$x^{2} \frac{d^{2} y}{d x^{2}} = n ((n + 1)) a x^{n + 1} + b x^{n + 1} = n (n + 1) y$

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If y=axn+1+bxn+1,thenx2d2ydx2 is equal to

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If $y = a x^{n + 1} + b x^{n + 1}, then x^{2} \frac{d^{2} y}{d x^{2}}$ is equal to