If y = (x^2 - 1)^m,then the (2m)^{th} differe | vedclass

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(B) Given $y = (x^2 - 1)^m$.Expanding this using the binomial theorem:$y = \binom{m}{0}(x^2)^m + \binom{m}{1}(x^2)^{m-1}(-1) + \dots + (-1)^m$.$y = x^

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$(2m)!$

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(B) Given $y = (x^2 - 1)^m$.Expanding this using the binomial theorem:$y = \binom{m}{0}(x^2)^m + \binom{m}{1}(x^2)^{m-1}(-1) + \dots + (-1)^m$.$y = x^{2m} - mx^{2m-2} + \dots + (-1)^m$.The $(2m)^{th}$ derivative of $x^{2m}$ is $(2m)!$.All other terms in the expansion are polynomials of degree less than $2m$.The derivative of any polynomial of degree $n$ with respect to $x$ taken $(n+1)$ times or more is $0$.Therefore,$\frac{d^{2m}y}{dx^{2m}} = \frac{d^{2m}}{dx^{2m}}(x^{2m}) - 0 + 0 - \dots = (2m)!$.

If $y = (x^2 - 1)^m$,then the $(2m)^{th}$ differential coefficient of $y$ is

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