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# 8.4: The Derivative

Difficulty Level: At Grade Created by: CK-12

## Learning objectives

• Demonstrate an understanding of the derivative of a function as a slope of the tangent line.
• Demonstrate an understanding of the derivative as an instantaneous rate of change.

## Introduction

The function f'(x) that we defined in previous section is so important that it has its own name: the derivative.

The Derivative
The function f' is defined by the formula
f(x)=limh0f(x+h)f(x)h$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}$
where f' is called the derivative of f with respect to x. The domain of f consists of all the values of x for which the limit exists.

Based on the discussion that we have had in previous section, the derivative f ' represents the slope of the tangent line at point x. Another way of interpreting it, the function y = f(x) has a derivative f ' whose value at x is the instantaneous rate of change of y with respect to point x.

Example 1: Find the derivative of f(x)=xx+1$f(x) = \frac{x} {x + 1}$.

Solution: We begin with the definition of the derivative,

f(x)=limh0f(x+h)f(x)h$f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}$

where

f(x)$f(x)$ =xx+1$= \frac{x} {x + 1}$
f(x+h)$f(x + h)$ =x+hx+h+1$= \frac{x + h} {x + h + 1}$

Substituting into the derivative formula,

f(x)$f'(x)$ =limh0[x+hx+h+1xx+1]h$= \lim_{h \rightarrow 0} \frac{\left [\frac{x + h} {x + h + 1} - \frac{x} {x + 1}\right ]}{h}$
=limh01h[x+hx+h+1xx+1]$= \lim_{h \rightarrow 0} \frac{1}{h} \left [\frac{x+h}{x+h+1}-\frac{x}{x+1}\right]$
=limh01h[(x+h)(x+1)x(x+h+1)(x+h+1)(x+1)]$= \lim_{h \rightarrow 0} \frac{1} {h} \left [\frac{(x + h) (x + 1) - x (x + h + 1)} {(x + h + 1) (x + 1)}\right ]$
=limh01h[x2+x+hx+hx2xhx(x+h+1)(x+1)]$= \lim_{h \rightarrow 0} \frac{1} {h} \left [\frac{x^2 + x + hx + h - x^2 - xh - x} {(x + h + 1) (x + 1)}\right ]$
=limh01h[h(x+h+1)(x+1)]$= \lim_{h \rightarrow 0} \frac{1} {h} \left [\frac{h} {(x + h + 1) (x + 1)}\right ]$
=limh01(x+h+1)(x+1)$= \lim_{h \rightarrow 0} \frac{1} {(x + h + 1) (x + 1)}$
=1(x+1)2$= \frac{1} {(x + 1)^2}$

Example 2: Find the derivative of f(x)=x$f(x) = \sqrt{x}$ and the equation of the tangent line at x0 = 1.

Solution:

Using the definition of the derivative,

f(x)$f'(x)$ =limh0f(x+h)f(x)h$= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}$
=limh0x+hxh$= \lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}} {h}$
=limh0x+hxh(x+h+xx+h+x)$= \lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}} {h} \left( \frac{\sqrt{x + h} + \sqrt{x}} {\sqrt{x + h} + \sqrt{x}} \right )$
=limh01hx+hxx+h+x$= \lim_{h \rightarrow 0} \frac{1} {h} \frac{x + h - x} {\sqrt{x + h} + \sqrt{x}}$
=limh01x+h+x$= \lim_{h \rightarrow 0} \frac{1} {\sqrt{x + h} + \sqrt{x}}$
=12x$= \frac{1} {2\sqrt{x}}$

Thus, the slope of the tangent line at x0 = 1 is

f(1)$f'(1)$ =121=12$= \frac {1}{2\sqrt{1}}=\frac {1}{2}$

For x0 = 1, we can find y0 by simply substituting into f(x):

f(x0)$f(x_0)$ y0$\equiv y_0$
f(1)$f(1)$ =1=1$= \sqrt{1}=1$
y0$y_0$ =1$= 1$

Thus the equation of the tangent line is

yy0$y-y_0$ =m(xx0)$= m(x-x_0)$
y1$y-1$ =12(x1)$= \frac {1}{2}(x-1)$
y$y$ =12x+12$= \frac {1}{2}x+\frac {1}{2}$

## Notations

Calculus, just like all branches of mathematics, is rich with notation. There are many ways to denote the derivative of a function y = f(x) beside the most popular one f'(x). Here they are:

f(x)$f'(x)$ dydx$\frac {dy}{dx}$ y$y'$ dfdx$\frac{df}{dx}$ df(x)dx$\frac {df(x)}{dx}$

In addition, when substituting the point x0 into the derivative we denote the substitution by one of the following notation

f(x0)$f'(x_0)$ dydxx=x0$\frac {dy}{dx}\big|_{x=x_0}$ dfdxx=x0$\frac{df}{dx}\big|_{x=x_0}$ df(x0)dx$\frac {df(x_0)}{dx}$

## Lesson Summary

In this lesson we formalized the slope of the tangent line by calling it the derivative. We learned that the derivative of a function is another function that gives us the instantaneous rate of change of the original function. We practiced finding the derivative of functions using the formal definition of the derivative that utilizes a limit.

## Points to Consider

1. As you do the review questions below, look for patterns in the derivatives. Do you think there might be a shortcut for finding the derivative of some functions?
2. How might knowledge of the derivative be useful?

## Review Questions

In problems 1-6, use the definition of the derivative to find f'(x) and then find the equation of the tangent line at x = x0.

1. f(x) = 6x2; x0 = 3.
2. f(x)=x+2;x0=8$f(x) = \sqrt {x+2}; x_0 = 8$
3. f(x) = 3x3 - 2; x0 = -1
4. f(x)=1x+2;x0=1$f(x) = \frac {1}{x+2}; x_0 = -1$
5. f(x) = ax2 - b , (where a and b are constants); x0 = b
6. f(x) = x1/3; x0 = 1.
7. Suppose that f has the property that f(x + y) = f(x) + f(y) + 3xy and limh0f(h)h=4$\lim_{h \to 0}\frac {f(h)}{h} = 4$. Find f(0) and f'(x).
8. Find dy/dx

## Date Created:

Feb 23, 2012

Jun 08, 2015
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