The n^{th} derivative of x^{n + 1} is | vedclass

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(A) Let $y = x^{n + 1}$.The first derivative is $y_1 = \frac{d}{dx}(x^{n+1}) = (n + 1)x^n$.The second derivative is $y_2 = \frac{d}{dx}((n + 1)x^n) =

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$(n + 1)!x$

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(A) Let $y = x^{n + 1}$.The first derivative is $y_1 = \frac{d}{dx}(x^{n+1}) = (n + 1)x^n$.The second derivative is $y_2 = \frac{d}{dx}((n + 1)x^n) = (n + 1)nx^{n-1}$.The third derivative is $y_3 = \frac{d}{dx}((n + 1)nx^{n-1}) = (n + 1)n(n-1)x^{n-2}$.Following this pattern,the $n^{th}$ derivative is given by $y_n = (n + 1)n(n-1)...(2)x^{n+1-n} = (n + 1)!x^1 = (n + 1)!x$.

The $n^{th}$ derivative of $x^{n + 1}$ is

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