1.10 Definition of Derivative

$\newenvironment {prompt}{}{} \newcommand {\tcolorboxenvironment }[2]{\BeforeBeginEnvironment {#1}{\begin {tcolorbox}[savedelimiter={#1},#2]}\AfterEndEnvironment {#1}{\end {tcolorbox}}} \newcommand {\tcbpkgprefix }[0]{} \newcommand {\bart }[0]{BART} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

In this section we learn the definition of derivative and we use it to solve the tangent line problem.

The Derivative

Given a function $y = f(x)$ , its derivative, denoted by $f'(x)$ , is a formula for the slope of its tangent lines. Conceptually, the derivative represents the instantaneous rate of change of the function. Other notations for the derivative include $y'$ and $\frac {dy}{dx}$ .

The interactive graph below shows a function and its tangent line at a point. You can drag the point along the curve and see how the tangent line changes. You can also zoom in on the curve to see that the tangent line approximates the curve locally.

The tangent line problem is to find the slope of the tangent line. To begin the discussion, we recall the formula for the slope of a line between two points, $(x_1, y_1)$ and $(x_2, y_2)$ : $m = \frac {\text {rise}}{\text {run}} = \frac {\Delta y}{\Delta x} = \frac {y_2 - y_1}{x_2 - x_1}.$

To find the slope of the tangent line to the graph of $y = f(x)$ at the point $(x, f(x))$ , we first compute the slope of the secant line connecting the point $(x, f(x))$ with the nearby point $(x+h, f(x+h))$ : $m_{\text {sec}} = \frac {f(x+h) - f(x)}{(x+h)-x}.$ Simplifying the denominator, we have $m_{\text {sec}} = \frac {f(x+h) - f(x)}{h}.$ This quantity is known as the difference quotient.

The slope of the tangent line, is obtained by letting $h \to 0$ which has the effect of moving the point $(x+h, f(x+h))$ towards the point $(x, f(x))$ .

This gives us the definition of the derivative:

Derivative The derivative of the function $f(x)$ , denoted by $f'(x)$ is defined by $f'(x) = \lim _{h \to 0} \frac {f(x+h)-f(x)}{h}.$ This definition is valid for all values of $x$ for which the limit exists.

The slope of the tangent line to the graph of the function $y=f(x)$ at the point $(x,f(x))$ is given by $m_{\text {tan}}= f'(x).$

Below is an interactive graph that shows both a tangent line (red) and a secant line (green). Consider the red dot to be the point $(x, f(x)$ . Move the red dot to a point of your choosing. The green dot represents the point $(x+h, f(x+h))$ . Move the green dot toward the red dot. Observe that when the green dot is very close to the red dot (i.e. $h$ is very small), the secant line becomes indistinguishable from the tangent line. If the green dot is to the left of the red dot, then $h$ is negative.

We now compute the derivative of several different functions.

Examples Using the Definition of the Derivative

We now compute the derivative of $x^2$ and interpret some of its values. Using the formula for $f'(x)$ , we have $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{(x+h)^2- x^2}{h}\\[5pt] &= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2)- x^2}{h} \\[5pt] &= \lim_{h \to 0} \frac{ (2xh + h^2)}{h}\\[5pt] &= \lim_{h \to 0} \frac{ h(2x + h)}{h}\\[5pt] &= \lim_{h \to 0} (2x + h)\\ &= 2x. \end{align*}$

Thus the derivative of $x^2$ is $2x$ and this tells use the slope of the tangent line at a given $x$ -value. The process of obtaining the derivative is called differentiation, and the mathematical symbol for the differentiation operator is $\frac {d}{dx}$ so that we can write: $\frac {d}{dx}\left ( x^2\right ) = 2x.$ This is essentially the notation used by German polymath, philosopher and co-discoverer of calculus, Gottfried Wilhelm Leibniz circa 1700.
Now let’s use the derivative to discuss tangent lines and their slopes.

At the point $(-3, 9)$ on the graph of the parabola, $f(x) = x^2$ , the slope of the tangent line is given by $f'(-3) = 2x\big |_{x=-3} = 2(-3) = -6.$ That is, we plug the x-coordinate, $x=-3$ , into the function $f'(x) = 2x$ .
The points $(0,0$ and $(2,4)$ are also on the parabola. At the point $(0,0)$ , the slope of the tangent line is $m_{\text {tan}}= f'(0) =2x\big |_{x=0}= 2(0) = 0,$ and at the point $(2, 4)$ the slope is $m_{\text {tan}}= f'(2) =2x\big |_{x=2}= 2(2) = 4.$

Let $f(x) = x^2 - 5x + 7$ . To find $f'(x)$ using the definition of derivative, we will need to compute $f(x+h)$ . To do this, we substitute the expression $(x+h)$ into the formula for $f(x)$ in the place of $x$ . In this example, we get, $\begin{align*} f(x+h) &= (x+h)^2 - 5(x+h) + 7 \\ &= x^2 + 2xh + h^2 - 5x -5h + 7. \end{align*}$

Now we use the definition:

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\[5pt] &= \lim_{h \to 0} \frac{(x^2 + 2xh + h^2 - 5x - 5h +7)- (x^2 - 5x + 7)}{h}\\[5pt] &= \lim_{h \to 0} \frac{ 2xh + h^2 - 5h }{h} \\[5pt] &= \lim_{h \to 0} \frac{ h(2x + h -5)}{h}\\[5pt] &= \lim_{h \to 0} (2x + h -5)\\ &= 2x-5. \end{align*}$

To recap, if $f(x) = x^2 - 5x + 7$ , then $f'(x) =2x-5$ , or $\frac {d}{dx} (x^2 - 5x + 7) = 2x-5.$

In general, if y = f(x) then $y' = \frac {dy}{dx} = \frac {d}{dx} f(x) = f'(x),$ are all valid ways to express the derivative. Furthermore, the following are valid ways to express the derivative evaluated at a point $(a, f(a))$ , as when finding the slope of a tangent line: $y'(a) = y' \big |_{x=a} = \frac {dy}{dx}\Big |_{x=a} = \frac {d}{dx} f(x)\Big |_{x=a} = f'(a).$

Use the fact that the derivative of $x^2 -5x + 7$ is $2x-5$ , i.e., $\frac {d}{dx}\left ( x^2 - 5x + 7 \right ) = 2x-5,$ to find the slope of the tangent line to the graph of $f(x) = x^2 - 5x + 7$ at the point where $x = 1$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(1) = (2x-5)\big |_{x=1}$

The slope is $\answer {-3}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = x^2 - 5x + 7 \ \ \text {at} \ \ x=1.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(1, f(1))$

Solve for $y$

The equation of the tangent line is $y = \answer {-3x+6}$

If $f(x) = x^3$ then in order to find $f'(x)$ we first compute $f(x+h)$ .

We have

$\begin{align*} f(x+h) &= (x+h)^3 \\ &= (x+h)(x+h)^2 \\ &= (x+h)(x^2 + 2xh + h^2)\\ &= x^3 + 2x^2h + xh^2 + x^2h + 2xh^2 + h^3\\ &= x^3 + 3x^2h + 3xh^2 + h^3. \end{align*}$

Now we use the definition:

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3)- x^3}{h}\\[5pt] &= \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}\\[5pt] &= \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h}\\[5pt] &= \lim_{h \to 0} (3x^2 + 3xh + h^2) \\ &= 3x^2. \end{align*}$

To recap, $\frac {d}{dx}x^3 = 3x^2.$ Try to notice a pattern in the form of the derivatives of $x^2$ and $x^3$ .

Use the fact that the derivative of $x^3$ is $3x^2$ , i.e., $\frac {d}{dx}\left ( x^3 \right ) = 3x^2,$ to find the slope of the tangent line to the graph of $f(x) = x^3$ at the point where $x = 2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(2)$

The slope is $\answer {12}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = x^3 \ \ \text {at} \ \ x=2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(2, f(2))$

Solve for $y$

The equation of the tangent line is $y = \answer {12x-16}$

If $f(x) = \sqrt x$ then $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{x+h}- \sqrt x}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{x+h}- \sqrt x}{h} \cdot \frac{\sqrt{x+h}+ \sqrt x}{\sqrt{x+h}+ \sqrt x} \\[5pt] &= \lim_{h \to 0} \frac{(x+h) - (x)}{h(\sqrt{x+h}+ \sqrt x)}\\[5pt] &= \lim_{h \to 0} \frac{h}{h(\sqrt{x+h}+ \sqrt x)}\\[5pt] &= \lim_{h \to 0} \frac{1}{\sqrt{x+h}+ \sqrt x}\\[5pt] &= \frac{1}{2 \sqrt x}. \end{align*}$

To recap, $\frac {d}{dx}\sqrt {x} = \frac {1}{2\sqrt {x}},$ which in the notation of exponents, where $\sqrt x = x^{1/2}$ , we can write $\frac {d}{dx}x^{1/2} = \frac 12 x^{-1/2}.$

Use the fact that the derivative of $\sqrt {x}$ is $\frac {1}{2\sqrt {x}}$ , i.e., $\frac {d}{dx}\sqrt {x} = \frac {1}{2\sqrt {x}},$ to find the slope of the tangent line to the graph of $f(x) = \sqrt {x}$ at the point where $x = 4$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(4)$

The slope is $\answer {1/4}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \sqrt {x} \ \ \text {at} \ \ x=4.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(4, f(4))$

Solve for $y$

The equation of the tangent line is $y = \answer {x/4 +1}$

If $f(x) = \sqrt {2x+1}$ then $\begin{align*} f(x+h) &= \sqrt{2(x+h)+1} \\ &= \sqrt{2x +2h+1}. \end{align*}$

Now for the derivative:

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{2x+2h+1}- \sqrt{2x+1}}{h}\\[5pt] &= \lim_{h \to 0} \frac{\sqrt{2x+2h+1}- \sqrt{2x+1}}{h} \cdot \frac{\sqrt{2x+2h+1}+ \sqrt{2x+1}}{\sqrt{2x+2h+1}+ \sqrt{2x+1}} \\[5pt] &= \lim_{h \to 0} \frac{(2x+2h+1) - (2x+1)}{h(\sqrt{2x+2h+1}+ \sqrt{2x+1})}\\[5pt] &= \lim_{h \to 0} \frac{2h}{h\sqrt{2x+2h+1}+ \sqrt{2x+1}}\\[5pt] &= \lim_{h \to 0} \frac{2}{\sqrt{2x+2h+1}+ \sqrt{2x+1}}\\[5pt] &= \frac{2}{\sqrt{2x+1}+ \sqrt{2x+1}}\\[5pt] &= \frac{2}{2\sqrt{2x+1}}\\[5pt] &= \displaystyle{\frac{1}{\sqrt{2x+1}}}. \end{align*}$

In the notation of Leibniz, $\frac {d}{dx}\sqrt {2x+1}= \frac {1}{\sqrt {2x+1}}.$

Use the fact that the derivative of $\sqrt {2x+1}$ is $\frac {1}{\sqrt {2x+1}}$ , i.e., $\frac {d}{dx}\sqrt {2x+1} = \frac {1}{\sqrt {2x+1}},$ to find the slope of the tangent line to the graph of $f(x) = \sqrt {2x+1}$ at the point where $x = 4$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(4)$

The slope is $\answer {1/3}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \sqrt {2x+1} \ \ \text {at} \ \ x=4.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(4, f(4))$

Solve for $y$

The equation of the tangent line is $y = \answer {x/3 + 5/3}$

If $f(x) = \displaystyle {\frac {1}{x}}$ then $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{1}{x+h}- \frac{1}{x}}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{(x) - (x+h)}{x(x+h)}}{h} \\[5pt] &= \lim_{h \to 0} \frac{-h}{x(x+h)}\cdot \frac{1}{h}\\[5pt] &= \lim_{h \to 0} \frac{-1}{x(x+h)} \\[5pt] &= -\frac{1}{x^2}. %\end{aligned}$ %\end{center} \end{align*}$

Using the differential operator, $d/dx$ , we can write: $\frac {d}{dx}\left ({\frac {1}{x}}\right ) = -\frac {1}{x^2}.$

Use the fact that the derivative of $\frac {1}{x}$ is $-\frac {1}{x^2}$ , i.e., $\frac {d}{dx}\left (\frac {1}{x}\right ) = -\frac {1}{x^2},$ to find the slope of the tangent line to the graph of $f(x) = \frac {1}{x}$ at the point where $x = 2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(2)$

The slope is $\answer {-1/4}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \frac {1}{x} \ \ \text {at} \ \ x=2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(2, f(2))$

Solve for $y$

The equation of the tangent line is $y = \answer {-x/4 + 1}$

If $f(x) = \displaystyle {\frac {2}{x+3}}$ then $\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{2}{x+h +3}- \frac{2}{x+3}}{h}\\[5pt] &= \lim_{h \to 0} \frac{\frac{2(x+3) - 2(x+h+3)}{(x+h+3)(x+3)}}{h} \\[5pt] &= \lim_{h \to 0} \frac{\frac{2x+6 - 2x-2h-6}{(x+h+3)(x+3)}}{h} \\[5pt] &= \lim_{h \to 0} \frac{-2h}{(x+h+3)(x+3)}\cdot \frac{1}{h}\\[5pt] &= \lim_{h \to 0} \frac{-2}{(x+h+3)(x+3)} \\[5pt] &= -\frac{2}{(x+3)^2}. \end{align*}$

To recap, the derivative of $\displaystyle {\frac {2}{x+3}}$ is $\displaystyle {-\frac {2}{(x+3)^2}}$ .

Use the fact that the derivative of $\frac {2}{x+3}$ is $-\frac {2}{(x+3)^2}$ , i.e., $\frac {d}{dx}\left (\frac {2}{x+3}\right ) = -\frac {2}{(x+3)^2},$ to find the slope of the tangent line to the graph of $f(x) = \frac {2}{x+3}$ at the point where $x = -2$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(-2)$

The slope is $\answer {-2}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \frac {2}{x+3} \ \ \text {at} \ \ x=-2.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(-2, f(-2))$

Solve for $y$

The equation of the tangent line is $y = \answer {-2x-2}$

The next two examples us the special limits:

$\lim _{h \to 0} \frac {\cos (h) -1}{h}=0 \;\;\mbox {and} \;\; \lim _{h \to 0}\frac {\sin (h)}{h} = 1.$

If $f(x) = \sin (x)$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h} \\[5pt] &= \lim _{h \to 0} \frac {\sin (x+h) - \sin (x)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin (x)\cos (h) + \cos (x)\sin (h) - \sin (x)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin (x)[\cos (h) -1] + \cos (x)\sin (h)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin (x)[\cos (h) -1]}{h} + \frac {\cos (x)\sin (h)}{h}\\[5pt] &= \sin (x) \cdot 0 + \cos (x) \cdot 1 \\[5pt] &= \cos (x). \end {aligned}$

This can be expressed simply as $\frac {d}{dx}\sin (x) = \cos (x),$ or in terms of any other independent variable, such as the common radian unit, $\theta$ : $\frac {d}{d\theta }\sin (\theta ) = \cos (\theta ).$

Use the fact that the derivative of $y = \sin (x)$ is $y' = \cos (x)$ , i.e., $\frac {d}{dx}\sin (x) = \cos (x),$ to find the slope of the tangent line to the graph of $f(x) = \sin (x)$ at the point where $x = 0$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(0)$

The slope is $\answer {1}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \sin (x) \ \ \text {at} \ \ x=0.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(0, f(0))$

Solve for $y$

The equation of the tangent line is $y = \answer {x}$

If $f(x) = \cos (x)$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h} \\[5pt] &= \lim _{h \to 0} \frac {\cos (x+h) - \cos (x)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos (x)\cos (h) - \sin (x)\sin (h) - \cos (x)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos (x)[\cos (h) -1] - \sin (x)\sin (h)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos (x)[\cos (h) -1]}{h} - \frac {\sin (x)\sin (h)}{h}\\[5pt] &= \cos (x) \cdot 0 - \sin (x)\cdot 1\\[5pt] &= -\sin (x).\\[5pt] \end {aligned}$

Using other variables, this can be written as $\frac {d}{dt} \cos (t) = -\sin (t) \\ \text {or}\\ \frac {d}{du} \cos (u) = -\sin (u).$

Use the fact that the derivative of $\cos (x)$ is $-\sin (x)$ , i.e., $\frac {d}{dx}\cos (x) = -\sin (x),$ to find the slope of the tangent line to the graph of $f(x) = \cos (x)$ at the point where $x = \pi$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(\pi )$

The slope is $\answer {0}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \cos (x) \ \ \text {at} \ \ x=\pi .$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(\pi , f(\pi ))$

Solve for $y$

The equation of the tangent line is $y = \answer {-1}$

The next example uses the special limit

$\lim _{h \to 0} \frac {e^h-1}{h} = 1.$

If $f(x) = e^x$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h} \\[5pt] &= \lim _{h \to 0}\frac {e^{x+h}-e^x}{h}\\[5pt] &= \lim _{h \to 0} \frac {e^x e^h-e^x}{h}\\[5pt] &= \lim _{h \to 0} \frac {e^x (e^h-1)}{h}\\[5pt] &= e^x \cdot 1 = e^x.\\[-5pt] \end {aligned}$

Here we encounter the fascinating fact that the rate of change of the natural exponential function is equal to the function itself, symbolized by the system $y=e^x\\ y' = e^x,$ so that $y' = y.$

Use the fact that the derivative of $e^x$ is itself, i.e., $\frac {d}{dx}e^x=e^x,$ to find the slope of the tangent line to the graph of $f(x) = e^x$ at the point where $x = 0$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(0)$

The slope is $\answer {1}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = e^x \ \ \text {at} \ \ x=0.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(0, f(0))$

Solve for $y$

The equation of the tangent line is $y = \answer {x+1}$

The next example uses the special limit

$\lim _{k \to 0} \frac {\ln (1 + k)}{k} = 1.$

If $f(x) = \ln (x)$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h}\\[5pt] &= \lim _{h \to 0}\frac {\ln (x+h)-\ln (x)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\ln (\frac {x+h}{x})}{h}\\[5pt] &= \lim _{h \to 0}\frac {\ln (1 + \frac {h}{x})}{h}\\[5pt] &= \lim _{k \to 0} \frac {\ln (1 + k)}{kx} \\[5pt] &= \frac {1}{x}. \end {aligned}$

Note that we made the substitution $k =\frac {h}{x}$ so that $h = kx$ and also note that $h\to 0$ is equivalent to $k\to 0$ . To recap, $\frac {d}{dx} \ln (x) = \frac {1}{x}$ which gives a memorable mathematical relationship between the transcendental natural logarithm function the rational reciprocal function.

Use the fact that the derivative of $\ln (x)$ is $\frac {1}{x}$ , i.e., $\frac {d}{dx}\ln (x) = \frac {1}{x},$ to find the slope of the tangent line to the graph of $f(x) = \ln (x)$ at the point where $x = 1$ .

The derivative gives the slope of the tangent line

The slope is $m_{\text {tan}} = f'(1)$

The slope is $\answer {1}$

Use the answer to the previous problem to find the equation of the tangent line to the graph of $f(x) = \ln (x) \ \ \text {at} \ \ x=1.$

Use the point slope form: $y-y_1 = m(x-x_1)$

The point $(x_1,y_1)$ is $(1, f(1))$

Solve for $y$

The equation of the tangent line is $y = \answer {x-1}$

Here are some detailed, lecture style video on the derivative: