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Question Number 163928 by mathlove last updated on 12/Jan/22

$$f\left(x\right)=\frac{2x^{100!}}{100!}+x^{100}+1$$$$find\frac{d^{100!}f\left(x\right)}{{dx}^{100!}}=?$$

Answered by mr W last updated on 12/Jan/22

$$generally:$$$$f\left(x\right)=x^m$$$$\frac{d^{n}f\left(x\right)}{{dx}^n}=m(m-1)(m-2)...(m-n+1)x^{m-n}$$$$\frac{d^{n}f\left(x\right)}{{dx}^n}=\frac{m!}{(m-n)!}{x}^{m-n}forn⩽m$$$$\frac{d^{n}f\left(x\right)}{{dx}^n}=0forn>m$$$$$$$$f\left(x\right)=\frac{2x^{100!}}{100!}+x^{100}+1$$$$\frac{d^{100!}f\left(x\right)}{{dx}^{100!}}=\frac{2}{100!}\times \frac{(100!)!}{(100!-100!)!}\times x^{100!-100!}+0+0$$$$\Rightarrow \frac{d^{100!}f\left(x\right)}{{dx}^{100!}}=\frac{2(100!)!}{100!}$$

Commented by mathlove last updated on 12/Jan/22

$$thanksmr$$

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