Question

Find the derivative of f of x is equal to 1 + x + x square + x cube + x ^ 50 at x is equal to 1

Answer 1

SOLUTION

GIVEN

$\sf{f(x) = 1 + x + {x}^{2} + {x}^{3} + {x}^{50} }$

TO DETERMINE

The derivative of f(x) at x = 1

EVALUATION

Here it is given that

$\sf{f(x) = 1 + x + {x}^{2} + {x}^{3} + {x}^{50} }$

Differentiating both sides with respect to x we get

$\sf{f'(x) = 1 +2 {x}^{} + 3{x}^{2} + 50{x}^{49} }$

Putting x = 1 in both sides we get

$\sf{f'(1) = 1 +2 \times 1 + 3 \times {(1)}^{2} + 50 \times {(1)}^{49} }$

$\sf{ \implies \: f'(1) = 1 + 2 + 3 + 50 }$

$\sf{ \implies \: f'(1) = 56 }$

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