For the function $f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\cdots+\dfrac{x^2}{2}+x+1$, prove that $f'(1)=100f'(0)$.

**OR**

For any constant $a$, find the derivative of $\dfrac{x^n-a^n}{x-a}$.

Question

For the function $f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\cdots+\dfrac{x^2}{2}+x+1$, prove that $f'(1)=100f'(0)$.

**OR**

For any constant $a$, find the derivative of $\dfrac{x^n-a^n}{x-a}$.

Accepted Answer

>

> [!TLDR]
> $f'(x)=x^{99}+x^{98}+\cdots+x+1$; evaluating at $x=1$ gives $100$ and at $x=0$ gives $1$, so $f'(1)=100f'(0)$.
>
> We solve the primary question (the OR alternative, on $\dfrac{x^n-a^n}{x-a}$, is not needed since this one is fully answerable).
>
> Given $f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\cdots+\dfrac{x^2}{2}+x+1$.
>
> Differentiate term by term, using $\dfrac{d}{dx}\left(\dfrac{x^k}{k}\right)=x^{k-1}$ for each $k=100,99,\dots,2,1$, and the derivative of the constant $1$ is $0$:
>
> $$f'(x) = x^{99}+x^{98}+\cdots+x^1+x^0 = \sum_{k=0}^{99}x^k$$
>
> (a sum of $100$ terms, powers $x^0$ through $x^{99}$).
>
> **At $x=1$:** every term equals $1^k=1$, and there are $100$ terms:
> …

Q.For the function $f(x)=\dfrac{x^{100}}{100}+\dfrac{x^{99}}{99}+\cdots+\dfrac{x^2}{2}+x+1$ , prove that $f'(1)=100f'(0)$ . OR For any constant $a$ , find the derivative of $\dfrac{x^n-a^n}{x-a}$ .

Concept understanding — Rules of Differentiation

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