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Question

# In Exercises , use the definition of the derivative to find $f^{\prime}(x)$.$f(x)=2 x^2+x-1$

Solution

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Function

$f(x)=2x^2+x-1$

is given. To find the derivative of $f$ by the definition, we need to find the following limit:

$\displaystyle\lim_{\Delta x\to0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}\text{.}$

We get:

\begin{align*} f'(x)&=\displaystyle\lim_{\Delta x\to0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}\\ &=\displaystyle\lim_{\Delta x\to0}{\frac{2(x+\Delta x)^2+(x+\Delta x)-1-(2x^2+x-1)}{\Delta x}}\\ &=\displaystyle\lim_{\Delta x\to0}{\frac{2(x^2+2x\Delta x+(\Delta x)^2)+x+\Delta x-1-2x^2-x+1}{\Delta x}}\\ &=\displaystyle\lim_{\Delta x\to0}{\frac{2x^2+4x\Delta x+2(\Delta x)^2+x+\Delta x-1-2x^2-x+1}{\Delta x}}\\ &=\displaystyle\lim_{\Delta x\to0}{\frac{4x\Delta x+2(\Delta x)^2+\Delta x}{\Delta x}}\\ &=\displaystyle\lim_{\Delta x\to0}{\frac{\Delta x(4x+2\Delta x+1)}{\Delta x}}\\ &=\displaystyle\lim_{\Delta x\to0}{4x+2\Delta x+1}\\ &=4x+2\cdot0+1\\ &=4x+1\text{.} \end{align*}

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