Limits & derivative

Calculus Level 1

Find the derivative of function f ( x ) = − 2 x 2 + 3 x + 1 f(x) = -2x^2 + 3x + 1 using the first principle.

4 x − 3 4x-3 − 4 x + 3 -4x+3

Ryan Shibu by Ryan Shibu , 16, India

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1 solution

James Watson
Jan 13, 2021

f ′ ( x ) ≡ lim ⁡ h → 0 f ( x + h ) − f ( x ) h f'(x) \equiv \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} Here, f ( x ) = − 2 x 2 + 3 x + 1 f(x) = -2x^2 + 3x + 1 so f ′ ( x ) = lim ⁡ h → 0 − 2 ( x + h ) 2 + 3 ( x + h ) + 1 − ( − 2 x 2 + 3 x + 1 ) h = lim ⁡ h → 0 − 2 x 2 − 4 x h − 2 h 2 + 3 x + 3 h + 1 + 2 x 2 − 3 x − 1 h = lim ⁡ h → 0 − 4 x h − 2 h 2 + 3 h h = lim ⁡ h → 0 − 4 x − 2 h + 3 = − 4 x + 3 \begin{aligned} f'(x) = \lim_{h\to 0} \frac{-2(x+h)^2 + 3(x+h) + 1 - (-2x^2 + 3x + 1)}{h} &= \lim_{h\to 0} \frac{-2x^2 - 4xh - 2h^2 + 3x+3h + 1 + 2x^2 - 3x - 1}{h} \\ &= \lim_{h\to 0} \frac{- 4xh - 2h^2 +3h}{h} \\ &= \lim_{h\to 0} -4x - 2h + 3 = \boxed{-4x+3} \end{aligned}

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