2.13: Definition of Derivative Examples
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In the last section, we saw the instantaneous rate of change, or derivative, of a function \(f(x)\) at a point \(x\) is given by
\(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
To use this in the formula \(f'(x) = \frac{f(x+h) - f(x)}{h}\), first we need to replace the \(f(x+h)\) part of the formula. This is the same as \(f(x)\) which is \(3x+5\), except we replace \(x\) with that \((x+h)\) in parantheses. Like the following. The colors are only to highlight the substitution of \(f(x+h)\) and \(f(x)\). We’ll drop the colors as soon as we need to combine expressions.
\[\begin{align*} f'(x) & = \lim_{h \to 0} \frac
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Now we continue to simplify and find the answer.
\[\begin{align*} f'(x) & = \lim_{h \to 0} \frac
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Here, we have \(f'(x) = 3\). That makes sense if you think about it: \(3x + 5\) is a line with slope \(3\)!
\[\begin{align*} f'(x) & = \lim_{h \to 0} \frac
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So what does this mean? Well, this means we double \(x\) to find the slope of the tangent line of \(f(x) =x^2\). So at \(x = 3\), the slope is \(6\), and at \(x = 1.2\), the slope is \(2.4\). ETC.