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  • For the function f x = x ^ 100/100 + x^99/99

5.   For the function f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1 Prove that f '(1) =100 f '(0).

Answers (1)

best_answer

f (x) = \frac{x^{100}}{100}+ \frac{x ^ {99}}{99} + ....+ \frac{x ^2 }{2}+ x+ 1

As we know, the property,

f'(x^n)=nx^{n-1}

applying that property we get 

f '(x) = 100\frac{x^{99}}{100}+ 99\frac{x ^ {98}}{99} + ....+ 2\frac{x }{2}+ 1+ 0

f '(x) = x^{99}+x^{98}+......x+1

Now.

f '(0) = 0^{99}+0^{98}+......0+1=1

f '(1) = 1^{99}+1^{98}+......1+1=100

So,

Here

 1\times 100=100

f'(0)\times 100=f'(1)

Hence Proved.

Posted by

Pankaj Sanodiya

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