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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives5 Marks
Question
For the function $\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+\dots+\frac{\text{x}^2}{2}+\text{x}+1.$Prove that $\text{f}'(1)=100\text{f}'(0).$
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Answer

We have,$\text{f}(\text{x})=\frac{\text{x}^{100}}{100}+\frac{\text{x}^{99}}{99}+\dots+\frac{\text{x}^2}{2}+\text{x}+1.$
Differentiating with respect to $\text{x},$ We get
$\text{f}'(\text{x})=\text{x}^{99}+\text{x}^{98}+\dots+\text{x}+1+0\dots(\text{i})$
From (i)
f'(1) = 1 + 1 + ...(100 times)
= 100
Again,
f'(0) = 0 + 0 + ... + 1
= 1
Now,
f'(1) = 100 = 100 $\times$ 1 = 100 $\times$ f'(0)
Hence'
f'(1) = 100f'(0)

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