2.3 The Basic Rules for Derivatives
2.3 The Basic Rules for Derivatives
In the next few sections, we’ll get the derivative rules that will let us find formulas for derivatives when our function comes to us as a formula. This is a very algebraic section, and you should get lots of practice. When you tell someone you have studied calculus, this is the one skill they will expect you to have.
Video Demonstration
Introduction to derivatives of formulas
© 2014 Eric Bancroft
Building Blocks
These are the simplest rules – rules for the basic functions. We won’t prove these rules; we’ll just use them. But first, let’s look at a few so that we can see they make sense.
Example 1
Find the derivative of [latex]y=f(x)=mx+b[/latex]
Answer:
This is a linear function, so its graph is its own tangent line! The slope of the tangent line, the derivative, is the slope of the line: [latex]f'(x)=m[/latex]
Linear Function Rule
Video Demonstration
Slope of a linear function
© 2014 Eric Bancroft
Example 2
Find the derivative of [latex]f(x)=135[/latex].
Answer: [latex]f'(x)=?[/latex]
[latex]f(x)=\overbrace{135}^{constant}=0[/latex]
Constant Function Rule
The derivative of a constant is zero.
[latex]\frac{d}{dx}\left(c\right)=0[/latex] if [latex]c[/latex] is a constant value
Video Demonstration
Slope of a constant function
© 2014 Eric Bancroft
Example 3
Find the derivative of [latex]f(x)=x^2[/latex].
Answer: [latex]f'(x)=?[/latex]
Recall the formal definition of the derivative: [latex]f'(x)=\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}.[/latex]
Using our function [latex]f(x)=x^2[/latex], [latex]f(x+h)=(x+h)^2=x^2+2xh+h^2[/latex].
Then
[latex]\begin{align*} f'(x) & = \lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}\\ & = \lim\limits_{h\to 0} \frac{x^2+2xh+h^2-x^2}{h}\\ & = \lim\limits_{h\to 0} \frac{2xh+h^2}{h}\\ & = \lim\limits_{h\to 0} \frac{h(2x+h)}{h}\\ & = \lim\limits_{h\to 0} (2x+h)\\ & = 2x \end{align*}[/latex]
From all that, we find that [latex]f'(x)=2x[/latex].
Luckily, there is a handy rule we use to skip using the limit:
Power Rule
The derivative of a power function is the original exponent multiplied by the original variable raised to the original exponent reduced by 1.
[latex]\frac{d}{dx}\left( x^n\right)=nx^{n-1}[/latex] where [latex]n[/latex] is a constant
Video Demonstration
Power rule – example 1; Power rule – example 2
© 2014 Eric Bancroft
Example 4
Find the derivative of [latex]g(x)=4x^3[/latex]
Answer: [latex]g'(x)=?[/latex]
Note that [latex]g(x)[/latex] is a constant multiple of a power function. As a result, we have to apply two rules: the constant multiple rule and the power rule.
[latex]g'(x)=\frac{d}{dx}\left(\overbrace{4\cdot \underbrace{x^3}_{power\ fn}}^{const.\ mult.}\right)=4\cdot 3x^{3-1}=12x^2[/latex]
Constant Multiple Rule
Derivative of a constant multiple of a function is the constant multiplied by the derivative of that function.
[latex]\frac{d}{dx}\left( c\cdot f\right)=k\cdot f'[/latex] if [latex]c[/latex] is a constant value
Video Demonstration
Constant multiples of functions
© 2014 Eric Bancroft
Here are all the basic rules in one place.
Derivative Rules: Building Blocks
In what follows, [latex]f[/latex] and [latex]g[/latex] are differentiable functions of [latex]x[/latex].
Constant Multiple Rule
[latex]\frac{d}{dx}\left( kf\right)=kf'[/latex]
Sum and Difference Rule
[latex]\frac{d}{dx}\left(f\pm g\right)=f' \pm g'[/latex]
Exponential Functions
[latex]\frac{d}{dx}\left(b^x\right)=\ln(b)\,b^x[/latex]
Logarithmic functions
[latex]\frac{d}{dx}\left(\log_b(x)\right)=\frac{1}{(\ln b)\,x}[/latex]
Power Rule
[latex]\frac{d}{dx}\left(x^n\right)=nx^{n-1}[/latex]
Special cases:
[latex]\frac{d}{dx}\left(k\right)=0 \quad \text{because }k=kx^0[/latex]
[latex]\frac{d}{dx}\left(x\right)=1 \quad \text{because } x=x^1[/latex]
[latex]\frac{d}{dx}\left(e^x\right)=e^x\quad \text{because }\ln e=1[/latex]
[latex]\frac{d}{dx}\left(\ln(x)\right)=\frac{1}{x}\quad \text{because }\ln e=1[/latex]
The sum, difference, and constant multiple rules combined with the power rule allow us to easily find the derivative of any polynomial.
Video Demonstration
Sum and difference of functions; Exponential function and natural log
© 2014 Eric Bancroft
Example 5
Find the derivative of [latex]p(x)=17x^{10}+13x^8-1.8x+1003[/latex].
Answer: [latex]p'(x)=?[/latex]
Analyzing operations and functions inside [latex]p(x)[/latex] we see that
[latex]p(x)=\overbrace{\overbrace{17\cdot \underbrace{x^{10}}_{power\ fn}}^{const.\ mult}+\overbrace{13\cdot \underbrace{x^8}_{power.\ fn}}^{const.\ mult.}-\overbrace{1.8\cdot \underbrace{x}_{lin.}}^{const.\ mult.}+\underbrace{1003}_{const.}}^{sum/difference}[/latex]
We can use this analysis to now apply the basic rules if derivatives to the function:
[latex]\begin{align*} \frac{d}{dx}\left(17x^{10}+13x^8-1.8x+1003 \right)& = 17\cdot 10x^{10-1}+13\cdot 8x^{8-1}-1.8\cdot 1+0\\\\ & = 170x^9+104x^7-1.8 \end{align*}[/latex]
Example 6
Find [latex]\frac{d}{dx}\left( 17x^2-33x+12 \right)[/latex].
Answer: [latex]\frac{d}{dx}\left( 17x^2-33x+12 \right)=?[/latex]
Analyzing the original function first, then applying the appropriate basic derivative rules gives us:
[latex]\frac{d}{dx}(\overbrace{\overbrace{17\cdot \underbrace{x^2}_{power\ fn.}}^{const.\ mult.}-\overbrace{33\underbrace{x}_{lin.}}^{const.\ mult.}+\underbrace{12}_{const.}}^{sum/difference} )=17(2x^{2-1})-33(1)+0=34x-33[/latex]
The power rule works even if the power is a negative number or a fraction. In order to apply it, first translate all roots and basic rational expressions into exponents:
Example 7
Find the derivative of [latex]y=3\sqrt{t}-\frac{4}{t^4}+5e^t[/latex].
Answer: [latex]y'=?[/latex]
The first step is translate roots and division by expressions involving the variable into power expressions: [latex]y=3\sqrt{t}-\frac{4}{t^4}+5e^t=3t^{1/2}-4t^{-4}+5e^t[/latex]
[latex]y=3\sqrt{t}-\frac{4}{t^4}+5e^t=\overbrace{\overbrace{3\cdot \underbrace{t^{1/2}}_{power\ fn.}}^{const.\ mult.}-\overbrace{4\cdot \underbrace{t^{-4}}_{power\ fn.}}^{const.\ mult.}+\overbrace{5\cdot \underbrace{e^t}_{nat.\ exp.\ fn.}}^{const.\ mult.}}^{sum/difference}[/latex]
Now we can take the derivative:
[latex]\begin{align*} \frac{d}{dt}(y)&=\frac{d}{dt}\left( 3t^{1/2}-4t^{-4}+5e^t \right)\\\\ &= 3\left(\frac{1}{2}\cdot t^{1/2-1}\right)-4\cdot \left(-4t^{-4-1}\right)+5\cdot \left(e^t\right) \\ & = \frac{3}{2}t^{-1/2}+16t^{-5}+5e^t\\\\ &= \frac{3}{2\sqrt{t}}+\frac{16}{t^5}+5e^t \end{align*}[/latex]
Note that, as far as the final answer goes, both of the last lines are correct and could be considered final. The last line is more appropriate in the sense that it brings the final answer back to the language of the original function, which was in terms of roots and non-negative exponents.
Another note: be careful when finding the derivatives with negative exponents.
We can immediately apply these rules to solve the problem we started the chapter with – finding a tangent line.
Example 8
Find the equation of the line tangent to [latex]g(t)=10-t^2[/latex] when [latex]t = 2[/latex].
Answer: tangent equation at [latex]t=2[/latex]?
Recall that the tangent of a function at a particular input value is the line sharing the point and the slope with the function at that input value. So the equation of the tangent to [latex]g(t)[/latex] at [latex]t=2[/latex] has the form [latex]y=mt+b[/latex], where [latex]m[/latex] is the derivative of [latex]g[/latex] at [latex]t=2[/latex] and the line passes through the point [latex](2, g(2))[/latex].
Since the slope of the tangent line is the value of the derivative at that point, we need to find [latex]g'(2)[/latex].
Since [latex]g[/latex] is a difference of a constant and a power function, we have that
[latex]g'(t)=0-2t^{2-1}=-2t[/latex]
and so [latex]m=g'(2)=-4.[/latex]
Therefore, the tangent equation is [latex]y=-4t+b[/latex] for some value [latex]b[/latex].
We now have to determine the value of [latex]b[/latex].
Since the tangent line passes the original function at [latex](2, g(2))[/latex], we have that
[latex]g(2)=g'(2)(2)+b\Rightarrow b=g(2)-g'(2)(2)=(10-2^2)-(-4)(2)=14[/latex]
Therefore, the equation of the tangent at [latex](2, g(2))[/latex] is [latex]y=-4t+14[/latex].
Graphing, and noting that [latex]g(2)=6[/latex], we can use this information to visualise the line that is tangent to the curve: link to graph
We can also use these rules to help us find the derivatives we need to interpret the behavior of a function.
Example 9
In a memory experiment, a researcher asks the subject to memorize as many words from a list as possible in 10 seconds. Recall is tested, then the subject is given 10 more seconds to study, and so on. Suppose the number of words remembered after [latex]t[/latex] seconds of studying could be modeled by [latex]W(t)=4t^{2/5}[/latex]. Find and interpret [latex]W'(20)[/latex].
Answer: [latex]W'(20)=?[/latex] Meaning?
Since [latex]W(t)[/latex] is a constant multiple of a power function, we have that
[latex]W'(t)=4\cdot \frac{2}{5}t^{2/5-1}=\frac{8}{5}t^{-3/5}[/latex]
Therefore,
[latex]W'(20)=\frac{8}{5}(20)^{-3/5}\approx 0.27[/latex]
Since [latex]W[/latex] is measured in words, and [latex]t[/latex] is in seconds, [latex]W'(20)\approx 0.27[/latex] means that, after 20 seconds of studying, the subject is learning about 0.27 more words per additional second of studying.
Example 10
If the revenue [latex]R[/latex] ($) from an electronics component used in a production of [latex]x[/latex] microchips can be determined by [latex]R(x)=x(125-3x)[/latex], determine [latex]R(10)[/latex] and [latex]R'(10)[/latex], and interpret the meaning of the results.
Answer: [latex]R(10)=?[/latex], [latex]R'(10)=?[/latex], meaning of [latex]R(10)[/latex]?, meaning of [latex]R'(10)[/latex]?
[latex]R(10)=?[/latex]
[latex]R(10)=10\cdot(125-3(10))=950[/latex]
[latex]R'(10)=?[/latex]
Need first [latex]R'(x)[/latex]. Let’s analyze [latex]R(x)[/latex] to determine which derivative rules to apply, and when:
[latex]R(x)=\overbrace{x\cdot \underbrace{(125-3x)}_{linear}}^{product\ -\ no\ rule\ (yet)\ for\ product!}\overset{rewrite}{=}\overbrace{\overbrace{125\cdot \underbrace{x}_{lin.}}^{const.\ mult.}-\overbrace{3\cdot \underbrace{x^2}_{power fn.}}^{const.\ mult}}^{difference}[/latex]
Therefore,
[latex]R'(x)=125\cdot 1-3\cdot 2x^{2-1}=125-6x[/latex]
and so
[latex]R'(10)=125-6(10)=65[/latex]
meaning of [latex]R(10)[/latex]?
[latex]R(10)=950[/latex] means that, at 10 microchips produced, the revenue from the electronics component is $950.
meaning of [latex]R'(10)[/latex]?
[latex]R'(10)=65[/latex] means that, at 10 microchips produced, the revenue from the electronics component is increasing by $65 per microchip.
Section Exercises
Work on the following exercises. Discuss your solutions with your peers and/or course instructor.
Section 2.3 Exercises – The Basic Rules for Derivatives
