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Differentiate the following function with respect to x.
For the function f(x)=$\dfrac{{{x}^{100}}}{100}+\dfrac{{{x}^{99}}}{99}+.....+\dfrac{{{x}^{2}}}{2}+x+1.$ Prove that f’(1)=100f’(0).

Answer
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Hint: Find differentiation of f(x) by using identity $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ then relate f’(1) and f’(0).

Complete step-by-step answer:
We have function given as;
f(x)=$\dfrac{{{x}^{100}}}{100}+\dfrac{{{x}^{99}}}{99}+.....+\dfrac{{{x}^{2}}}{2}+x+1..........\left( 1 \right)$
As, we have to prove
f ’(1)=100 f ’(0), so we need to calculate the first differentiation of f (x) or f ’(x).
Let us differentiate equation (1) with respect to x
$\dfrac{d}{dx}f(x)=f'(x)=\dfrac{d}{dx}\left( \dfrac{{{x}^{100}}}{100}+\dfrac{{{x}^{99}}}{99}+.....\dfrac{{{x}^{2}}}{2}+x+1 \right)$
As we have property of differentiation that
$\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right)+{{f}_{2}}\left( x \right)+.......{{f}_{n}}\left( x \right) \right)=\dfrac{d}{dx}\left( \left( {{f}_{1}}\left( x \right) \right)+\dfrac{d}{dx}\left( {{f}_{2}}\left( x \right) \right)+.......\dfrac{d}{dx}\left( {{f}_{n}}\left( x \right) \right) \right)$
Using the above property, we can write f ’(x) as
$f'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{{{x}^{100}}}{100} \right)+\dfrac{d}{dx}\left( \dfrac{{{x}^{99}}}{99} \right)+\dfrac{d}{dx}\left( \dfrac{{{x}^{98}}}{98} \right)+......\dfrac{d}{dx}\left( \dfrac{{{x}^{2}}}{2} \right)+\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( 1 \right)$
We have property of differentiation as
$\dfrac{d}{dx}\left( cf\left( x \right) \right)=c\dfrac{d}{dx}f\left( x \right)$ where c = constant
$\dfrac{d}{dx}\left( \text{constant} \right)=0$
Applying both the above properties with equation f ‘(x), we get;
$f'\left( x \right)=\dfrac{1}{100}\dfrac{d}{dx}\left( {{x}^{100}} \right)+\dfrac{1}{99}\dfrac{d}{dx}\left( {{x}^{99}} \right)+.....\dfrac{1}{2}\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( x \right)+0........\left( 2 \right)$
Now, we know differentiation of ${{x}^{n}}\text{ is }n{{x}^{n-1}}\text{ or }\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Using the above identity to the equation (2), we get;
$f'\left( x \right)=\dfrac{100}{100}{{x}^{99}}+\dfrac{99}{99}{{x}^{98}}+\dfrac{98}{98}{{x}^{97}}.....\dfrac{2}{2}{{x}^{1}}+1$
On simplifying the above equation, we get;
$f'\left( x \right)={{x}^{99}}+{{x}^{98}}+{{x}^{97}}.....{{x}^{1}}+1............\left( 3 \right)$
Now, coming to the question, we have to prove that
f ‘ (1) = 100 f ‘ (0)……………….(4)
Here, LHS past can be written from the equation (3) by just putting x=1 to both sides, we get;
$\begin{align}
  & f'\left( 1 \right)=\left( {{1}^{99}} \right)+\left( {{1}^{98}} \right)+\left( {{1}^{97}} \right)+.....1+1 \\
 & f'\left( 1 \right)=\dfrac{\left( 1+1+1+....1 \right)}{100times} \\
 & f'\left( 1 \right)=100 \\
\end{align}$
For RHS part i.e. 100 f ‘ (0), we can get f ‘ (0) by just putting x = 0 in equation (3)
$\begin{align}
  & f'\left( 0 \right)={{\left( 0 \right)}^{99}}+{{\left( 0 \right)}^{98}}+{{\left( 0 \right)}^{97}}.....0+1 \\
 & f'\left( 0 \right)=1 \\
\end{align}$
RHS part is given as 100 f ‘ (0); Therefore,
RHS = f ‘ (0) = 100……………(6)
From equation (5) and (6), we get that the right hand side of both the equations are equal which means left should also be equal. Hence,
f '(1) =100 f ‘ (0)
Hence, Proved.
Note: One can go wrong while counting the number of 1’s in equation (5). We need to relate it with the powers given in equation (3) i.e. from 99 to 0 (powers) which in total is 100. Hence, the number of terms in f ‘ (x) will be 100.
One can go wrong when he/she tries to calculate first f (0) and f (1) from the given function i.e.
f (0) = 1 and
$f\left( 1 \right)=\dfrac{1}{100}+\dfrac{1}{99}+....1$
And now differentiate the above f (0) and f (1) and will get f ‘ (0) = 0 and f ‘ (1) = 0 which is wrong.
Hence, if we need to find f ‘ (constant) if f (x) is given, then first we have to find f ‘ (x), after that put x = constant get f ‘ (constant).