The Concept of the Derivative#

The Basic Definitions#

Slope of a Tangent Line

The slope of the tangent line to the graph of \(f(x)\) at the point \(P(x, f(x))\) is given by

\[\lim_{h\to 0} \frac{f(x+h) - f(x)}{h},\]

if the limit exists.

Average and Instantaneous Rate of Change

The average rate of change of \(f(x)\) over the interval \((x, x+h)\) or slope of the secant line to the graph of \(f(x)\) through the points \((x,f(x))\) and \((x+h, f(x+h))\) is

\[ \frac{f(x+h) - f(x)}{h}.\]

The instantaneous rate of change of \(f(x)\) at \(x\) or slope of the tangent line to the graph of \(f(x)\) at \((x,f(x))\) is

\[\lim_{h\to 0} \frac{f(x+h) - f(x)}{h},\]

if the limit exists.

The Limit Definition of the Derivative

The derivative of a function \(f(x)\) with respect to \(x\) is the function \(f'(x)\) which is defined as follows

\[f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h},\]

and the domain of \(f'(x)\) is the set of all \(x\) where the limit exists.

Differentiability and Continuity

If a function is differentiable at \(x=a\), then it is continuous at \(x=a\). If a function is continuous at \(x=a\), then it is not necessarily differentiable at \(x=a\).

Computing Derivatives Using the Limit Definition#

Example 1#

Find the slope of the tangent line to the function

\[f(x)=3x^2+12\]

at \(x=5\) using the limit definition of the derivative.

Step 1:   Write down the limit definition of a derivative.
\[ \boxed{f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}} \]

Remember: \(f(x+h)\) means that we take \(f(x)\) and replace \(x\) with \((x+h)\). For example, if \(f(x) = x^2\), then \(f(x+h)=(x+h)^2=x^2+2xh+h^2.\)

Step 2:   Plug   \(f(x+h)\)   and   \(f(x)\)   into definition.

Plug \(f(x+h)\) and \(f(x)\) into the limit definition of the derivative. Using brackets will help avoid errors from forgetting to distribute the negative sign:

\[ f'(x)=\lim_{h \to 0} \frac{[3(x+h)^2+12]-[3x^2+12]}{h}\]
Step 3:   FOIL and Simplify
\[\begin{align*} f'(x) &=\lim_{h \to 0} \frac{[3(x+h)^2+12]-[3x^2+12]}{h}\\ \\ &=\lim_{h \to 0} \frac{[3(x^2+2xh+h^2)+12]-[3x^2+12]}{h} && \hbox{FOIL $(x+h)^2$}\\ \\ &=\lim_{h \to 0} \frac{3x^2+6xh+3h^2+12-3x^2-12}{h} && \hbox{Distribute the '3' and the minus sign}\\ \\ &=\lim_{h \to 0} \frac{\cancel{3x^2}+6xh+3h^2+\cancel{12}-\cancel{3x^2}-\cancel{12}}{h} && \hbox{Simplify}\\ \\ &=\lim_{h \to 0} \frac{6xh+3h^2}{h} \end{align*}\]
Step 4:   Factor out   \(h\)   and cancel.

Factor out an \(h\) in the numerator and cancel it with the factor of \(h\) in the denominator.

\[f'(x)=\lim_{h \to 0} \frac{6xh+3h^2}{h}=\lim_{h \to 0} \frac{h(6x+3h)}{h}=\lim_{h \to 0} 6x+3h\]
Step 5:   Evaluate the limit.
\[f'(x)=\lim_{h \to 0} 6x+3h=6x+3(0)=6x\]
Step 6:   Plug   \(x=5\)   into evaluated limit.

We have found that \(f'(x) = 6x\) is the derivative of our function and the general form of the slope of the tangent line. All that’s left for us to do is to plug in \(x=5\). Therefore, the slope of the tangent line when \(x=5\) is \(30\).

Example 2#

Find the derivative of the function

\[f(x)=3\sqrt{x}\]

using the limit definition of the derivative.

Step 1:   Write down the limit definition of a derivative.
\[ \boxed{f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}} \]
Step 2:   Plug   \(f(x+h)\)   and   \(f(x)\)   into definition.

Plug \(f(x+h)\) and \(f(x)\) into the limit definition of the derivative.

\[\begin{align*} f'(x) = \lim_{h \to 0} \frac{3\sqrt{x+h}-3\sqrt{x}}{h} = \lim_{h \to 0} \frac{3(\sqrt{x+h}-\sqrt{x})}{h} \end{align*}\]
Step 3:   Rationalize, FOIL, and Simplify.
\[\begin{align*} f'(x) &= \lim_{h \to 0} \frac{3(\sqrt{x+h}-\sqrt{x})}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} && \hbox{Rationalize} \\ \\ &= \lim_{h \to 0} \frac{3(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})} \\ \\ &= \lim_{h \to 0} \frac{3(\sqrt{x+h}\sqrt{x+h} + \sqrt{x+h}\sqrt{x}-\sqrt{x}\sqrt{x+h} - \sqrt{x}\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})} && \hbox{FOIL} \\ \\ &= \lim_{h \to 0} \frac{3(x+h + \cancel{\sqrt{x+h}\sqrt{x}}-\cancel{\sqrt{x}\sqrt{x+h}} - x)}{h(\sqrt{x+h}+\sqrt{x})} && \hbox{Simplify} \\ \\ &= \lim_{h \to 0} \frac{3(\cancel{x}+h-\cancel{x})}{h(\sqrt{x+h}+\sqrt{x})} \\ \\ &= \lim_{h \to 0} \frac{3 h}{h(\sqrt{x+h}+\sqrt{x})} \end{align*}\]

Notice how making use of the formula \((\sqrt{A} - \sqrt{B})(\sqrt{A} + \sqrt{B}) = A - B\) can help eliminate some of the above computations.

Step 4:   Cancel the   \(h\)   in the numerator with the   \(h\)   in the denominator.
\[f'(x) = \lim_{h \to 0} \frac{3 \cancel{h}}{\cancel{h}(\sqrt{x+h}+\sqrt{x})} = \lim_{h \to 0} \frac{3}{\sqrt{x+h}+\sqrt{x}} \]
Step 5:   Evaluate the limit.
\[f'(x) = \lim_{h \to 0} \frac{3}{\sqrt{x+h}+\sqrt{x}} = \frac{3}{\sqrt{x + 0}+\sqrt{x}} = \frac{3}{\sqrt{x}+\sqrt{x}} = \frac{3}{2\sqrt{x}}\]