Find the derivative of f(x) = 1 + x + x^{2} + | vedclass

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(A) The given function is $f(x) = 1 + x + x^{2} + x^{3} + \dots + x^{50}$.Taking the derivative with respect to $x$,we get $f'(x) = \frac{d}{dx}(1) +

Vedclass · Accepted Answer

$1275$

Vedclass · Answer

(A) The given function is $f(x) = 1 + x + x^{2} + x^{3} + \dots + x^{50}$.Taking the derivative with respect to $x$,we get $f'(x) = \frac{d}{dx}(1) + \frac{d}{dx}(x) + \frac{d}{dx}(x^{2}) + \dots + \frac{d}{dx}(x^{50})$.Using the power rule $\frac{d}{dx}(x^{n}) = nx^{n-1}$,we have $f'(x) = 0 + 1 + 2x + 3x^{2} + \dots + 50x^{49}$.At $x = 1$,the value is $f'(1) = 1 + 2(1) + 3(1)^{2} + \dots + 50(1)^{49}$.This simplifies to the sum of the first $50$ natural numbers: $1 + 2 + 3 + \dots + 50$.The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$.For $n = 50$,the sum is $\frac{50 	imes 51}{2} = 25 	imes 51 = 1275$.

Find the derivative of $f(x) = 1 + x + x^{2} + x^{3} + \dots + x^{50}$ at $x = 1$.

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