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
Verified
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).
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