If f(x) = (x^2 - 1)^7, then f^{(14)}(x) is eq | vedclass

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(D) Given, $f(x) = (x^2 - 1)^7$. Using the binomial expansion, $f(x) = \sum_{k=0}^{7} \binom{7}{k} (x^2)^{7-k} (-1)^k$. The highest power of $x$ in th

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$14!$

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(D) Given, $f(x) = (x^2 - 1)^7$. Using the binomial expansion, $f(x) = \sum_{k=0}^{7} \binom{7}{k} (x^2)^{7-k} (-1)^k$. The highest power of $x$ in this expansion is $x^{14}$, which occurs when $k=0$. The term containing $x^{14}$ is $\binom{7}{0} (x^2)^7 (-1)^0 = 1 \cdot x^{14} \cdot 1 = x^{14}$. The $n$-th derivative of $x^n$ is $n!$. Therefore, the $14$-th derivative of $x^{14}$ is $14!$. Since all other terms in the expansion have powers of $x$ less than $14$, their $14$-th derivatives are $0$. Thus, $f^{(14)}(x) = 14!$.

If $f(x) = (x^2 - 1)^7$, then $f^{(14)}(x)$ is equal to

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