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Find the derivative of the function f(x) = 2x^2-x using the limit definition of a derivative.

Question:

Find the derivative of the function {eq}f(x) = 2x^2-x {/eq} using the limit definition of a derivative.

Answer and Explanation:

Consider the function {eq}f(x) = 2x^2 - x. {/eq} The first derivative is defined as the limit:

{eq}\begin{eqnarray*} f'(x) &=& \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h} \\ &=& \lim_{h \to 0} \dfrac{( 2 (x+h)^2 - (x+h)) -(2x^2 - x)}{h} \\ &=& \lim_{h \to 0} \dfrac{ 2 (x^2 + 2x h + h^2) - x - h - 2x^2 + x}{h} \\ &=& \lim_{h \to 0} \dfrac{ 2x^2 + 4x h + 2h^2 - x - h - 2x^2 + x}{h} \\ &=& \lim_{h \to 0} \dfrac{4x h + 2h^2 - h}{h} \\ &=& \lim_{h \to 0} 4x - 1 + 2h = 4x - 1 \end{eqnarray*} {/eq}


Learn more about this topic:

How to Determine if a Limit Does Not Exist

from AP Calculus AB: Exam Prep

Chapter 4 / Lesson 9
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