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Question Number 178250 by zaheen last updated on 14/Oct/22

$$howisthesolutionofthisqustion$$$$f\left(x\right)=x(x-1)(x-2)(x-3)(x-4)\cdot .......\cdot (x-100)$$$$f^{{}^{\prime }}\left(x\right)=?f^{\prime }\left(1\right)=?$$$$$$

Answered by CElcedricjunior last updated on 14/Oct/22

$$f\left(x\right)=\mathit{x}(\mathit{x}-1)(\mathit{x}-2)(\mathit{x}-3)....(\mathit{x}-\mathit{n})$$$$\mathit{f}\left(\mathit{x}\right)=\underset{\mathit{k}=1}{\prod ^{100}}\mathit{x}(\mathit{x}-\mathit{k})=\mathit{x}(\mathit{x}-100)!$$$${\mathit{f}}^{\prime }\left(\mathit{x}\right)=\frac{\mathbf{x}(\mathbf{x}-1)(\mathbf{x}-2)....(\mathbf{x}-\mathbf{n})}{\mathbf{x}}+$$$$\frac{\mathbf{x}(\mathbf{x}-1)(\mathbf{x}-2)...(\mathbf{x}-\mathbf{n})}{\mathbf{x}-1}+\frac{\mathbf{x}(\mathbf{x}-1)...(\mathbf{x}-\mathbf{n})}{\mathbf{x}-2}$$$$+.....+\frac{\mathbf{x}(\mathbf{x}-1)(\mathbf{x}-2)...(\mathbf{x}-100)}{\mathbf{x}-100}$$$${\mathit{f}}^{\prime }\left(\mathit{x}\right)=\underset{\mathit{k}=0}{\sum ^{100}}\frac{\mathbf{x}(\mathbf{x}-100)!}{\mathbf{x}-\mathbf{k}}$$$${\mathbf{f}}^{\prime }\left(1\right)=1\times (-1)\times (-2)\times .....(1-100)$$$${\mathit{f}}^{\prime }\left(1\right)=-99!$$$$............lecelebrecedricjunior.......$$

Answered by Sheshdevsahu last updated on 14/Oct/22

$$f^{\prime }\left(x\right)=\{1.(x-1)(x-2).....(x-100)\}+\{x.1.(x-2)....(x-100)\}+.......\{x(x-1)(x-2)....(x-99).1\}$$$$f^{\prime }\left(1\right)=\left(0\right)+\{1.1(1-2)(1-3)....(1-100)\}+.....\left(0\right)....+\left(0\right)$$$$f^{\prime }\left(1\right)=1(-1)(-2)(-3).....(-98)(-99)$$$$f^{\prime }\left(1\right)=-99!\left(Ans\right)$$$$$$$$f^{\prime }\left(x\right)=\frac{x!}{x}+\frac{x!}{x-1}+.......\frac{x!}{x-100}$$$$$$

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