8.10: Derivatives of Sums and Differences
Juan has been out playing with his model rocket all afternoon. Partway through the day, he started taking videos of the flights using his cell phone. Watching the video, he notices that the rockets actually seem to be getting faster after the launch instead of starting off at full speed and slowing down due to gravity.
Juan figures it is reasonable to assume it takes a bit for the engines to get the rocket up to full speed, but the acceleration seems to continue past when he figures that would continue.
After considering for a while, he wonders if the decreased mass of the rocket as it burns fuel might be the cause, assuming he knows the force generated by the engines and the starting and ending weight of the rocket, is there a way he could conjecture whether the increased acceleration might be a result of the decreased mass?
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Guidance
Derivatives of Sums and Differences
Theorem: If
and
In simpler notation


The Product Rule
Theorem: (The Product Rule) If
In a simpler notation
In words, The derivative of the product of two functions is equal to the first function times the derivative of the second plus the second function times the derivatives of the derivative of the first.
Keep in mind that 

Example A
Find the derivative of:


f(x)=3x2+2x

Solution
Use the power rule to help:









Example B
Find the derivative:
Solution
Again use the power rule to help:







Example C
Find
Solution
There two methods to solve this problem. One is to multiply the product and then use the derivative of the sum rule. The second is to directly use the product rule. Either rule will produce the same answer. We begin with the sum rule.
y 
= 


= 
 Taking the derivative of the sum yields





 Now we use the product rule.









 Which is the same answer.
Concept question wrapup Yes, he can make the conjecture. Assuming that the force is equal to the change in mass times velocity (momentum) over change in time, then using the power rule and simplifying, he can discover that the acceleration of the rocket is equal to the force minus the velocity multiplied by change in mass over time, all divided by mass, in mathematics this looks like:
Looking at the upperright portion of the equation, we can see that as mass decreases, the fraction Looks like Juan was right. 

Vocabulary
The product rule states that the derivative of the product of two functions equals the first function
A differentiable function is one which has a derivative that can be calculated.
Guided Practice
Questions
Find the Derivative
1) Given:

What is
dtdx whenx=0
2) What is the derivative of
3) Given

What is
dydx
4) Given

Find
c(−2) assumingy′(−2)=11 .
5) What is \begin{align*}\frac{d}{dx}[(5x)(cos x)]?\end{align*}
Solutions
1) By the difference rule: \begin{align*}(x  1)' = (x)'  (1)' = 0\end{align*}

\begin{align*}x' = 1\end{align*}
x′=1 ..... By the power rule 
\begin{align*}1' = 0\end{align*}
1′=0 ..... The derivative of a constant = 0  So when we evaluate this at x = 0, we get 1, since 1  0 = 1
2) We'll use the difference rule

First, expand \begin{align*}(x 1)(x + 1) \to x^2 2x 1.\end{align*}
(−x−1)(x+1)→−x2−2x−1. 
By the difference rule: \begin{align*}(x^2 2x 1)' = (x^2)' (2x)' (1)' = 2x 2\end{align*}
(−x2−2x−1)′=(−x2)′−(2x)′−(1)′=−2x−2
3) We'll use the difference and power rules:

\begin{align*}\frac{d}{dx}(\pi x^{0.54} + 6x^4) =\end{align*}
ddx(−πx−0.54+6x4)= 
\begin{align*}\frac{d}{dx}(\pi x^{0.54}) + \frac{d}{dx}(6x^4)\end{align*}
ddx(−πx−0.54)+ddx(6x4) ..... By the difference rule 
\begin{align*}\to 0.54 \pi x^{1.54} + 24x^3\end{align*}
→0.54πx−1.54+24x3 ..... By the power rule
4) We'll apply the product rule:

\begin{align*}(yc)' = y'c + yc'\end{align*}
(yc)′=y′c+yc′ 
\begin{align*}(yc)'(2) = y'(2)c(2) + y(2)c'(2)\end{align*}
(yc)′(−2)=y′(−2)c(−2)+y(−2)c′(−2) ..... By the product rule 
\begin{align*}132 = 11c(2) + (0)c'(2)\end{align*}
132=11c(−2)+(0)c′(−2) ..... Substitute the given values 
\begin{align*}132 = 11c(2)\end{align*}
132=11c(−2) ..... Because \begin{align*}(0)c'(2) = 0\end{align*}(0)c′(−2)=0 
\begin{align*}12 = c(2)\end{align*}
12=c(−2) ..... Simplify
5) We'll use the product rule:

\begin{align*}(pq)' = p'q + pq'\end{align*}
(pq)′=p′q+pq′ . 
\begin{align*}p(x) = 5x \to p'(x) = 5\end{align*}
p(x)=−5x→p′(x)=−5 .... By the power rule 
\begin{align*}q(x) = cos x \to q'(x) = sin x\end{align*}
q(x)=cosx→q′(x)=−sinx ..... By the power rule and simplifying 
So we get \begin{align*}[(5x)(cos x)]' = (5)(cos x) + (5x)(sin x)\end{align*}
[(−5x)(cosx)]′=(−5)(cosx)+(−5x)(−sinx) 
\begin{align*}= 5cos x + 5xsin x\end{align*}
=−5cosx+5xsinx
Practice
Find the Derivative using the sum/difference rule

\begin{align*}y = \frac{1} {2} (x^3  2x^2 + 1)\end{align*}
y=12(x3−2x2+1) 
\begin{align*}y = \sqrt{2} x^3 \frac{1} {\sqrt{2}}x^2 + 2x + \sqrt{2}\end{align*}
y=2√x3−12√x2+2x+2√ 
\begin{align*}y = a^2  b^2 + x^2  a  b + x\end{align*}
y=a2−b2+x2−a−b+x (where a, b are constants) 
\begin{align*}y = x^{3} + \frac{1} {x^7}\end{align*}
y=x−3+1x7  \begin{align*}y = \sqrt{x} + \frac{1} {\sqrt{x}} \end{align*}
 \begin{align*}f(x) = (3x + 4)^2\end{align*}
 \begin{align*}f(x) = 0.93x^{10} + (\pi^3x)^{\frac{5}{12}}\end{align*}
 What is \begin{align*}\frac{d}{dx} (2x + 1)^2\end{align*}
 Given: \begin{align*}a(x) = (5x + 3)^2\end{align*} What is \begin{align*}\frac{dy}{dx}\end{align*}
 \begin{align*} v(x) = 3x^3 + 5x^2  2x  3\end{align*} What is \begin{align*}v'(0)\end{align*}
Find the Derivative using the product rule
 \begin{align*}y = (x^3  3x^2 + x) \cdot (2x^3 + 7x^4)\end{align*}
 \begin{align*}y = \left (\frac{1} {x} + \frac{1} {x^2}\right ) (3x^4  7) \end{align*}
 What is \begin{align*}[(3x^2 + x + 4)(3x  3)]\end{align*}
 \begin{align*}v(x) = (3x  3)(cos x)\end{align*}
 Given: \begin{align*}k(2) = 0\end{align*} \begin{align*}k'(2) = 18\end{align*} Find \begin{align*}r(2)\end{align*} when \begin{align*}(kr)'(2) = 54\end{align*}
 Given \begin{align*}g(x) = (4x^2  4x  5)(3x  3)\end{align*} Find \begin{align*}g'(2)\end{align*}
 Find \begin{align*}\frac{d}{dx}[(4x + 3)(sin x)\end{align*}
 Find \begin{align*}\frac{d}{dx}[(x^2  3) (2x^2 + 4x  1)]\end{align*}
 Given \begin{align*}t(1) = 0\end{align*} \begin{align*}t'(1) = 17\end{align*} Find \begin{align*}a(1)\end{align*} when \begin{align*}(ta)'(1) = 272\end{align*}
 Given \begin{align*}d(x) =(2x^2 + 3x  1)(2x + 1)\end{align*} Find \begin{align*}d'(1)\end{align*}
derivative
The derivative of a function is the slope of the line tangent to the function at a given point on the graph. Notations for derivative include , , , and \frac{df(x)}{dx}.differentiable
A differentiable function is a function that has a derivative that can be calculated.product rule
In calculus, the product rule states that the derivative of the product of two functions equals the first function times the derivative of the second function, added to the second function times the derivative of the first function.theorem
A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.Image Attributions
Description
Learning Objectives
Here you will learn about finding the derivative of functions that are added, subtracted, or multiplied. You will also explore another theorem: The Product Rule.