<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Derivatives of Sums and Differences

## Rules for derivatives when functions are added, subtracted, or multiplied.

0%
Progress
Practice Derivatives of Sums and Differences
Progress
0%
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?

Embedded Video:

### Guidance

Derivatives of Sums and Differences

Theorem: If f\begin{align*}f\end{align*} and g\begin{align*}g\end{align*} are two differentiable functions at x\begin{align*}x\end{align*} then

ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)]\begin{align*}\frac {d}{dx}\left [{f(x)+g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]+\frac {d}{dx}\left [{g(x)} \right ]\end{align*}

and

ddx[f(x)g(x)]=ddx[f(x)]ddx[g(x)]\begin{align*}\frac {d}{dx}\left [{f(x)-g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]- \frac {d}{dx}\left [{g(x)} \right ]\end{align*}

In simpler notation

(f±g)=f±g\begin{align*}(f \pm g)'= f'\pm g'\end{align*}.

The Product Rule

Theorem: (The Product Rule) If f\begin{align*}f\end{align*} and g\begin{align*}g\end{align*} are differentiable at x\begin{align*}x\end{align*}, then

ddx[f(x)g(x)]=f(x)ddxg(x)+g(x)ddxf(x)\begin{align*}\frac {d}{dx}\left [{f(x)\cdot g(x)} \right ]= f(x) \frac {d}{dx}g(x) + g(x)\frac {d}{dx}f(x)\end{align*}

In a simpler notation

(fg)=fg+gf\begin{align*}(f\cdot g)'=f\cdot g'+g\cdot f'\end{align*}

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 (fg)f+g\begin{align*}(f\cdot g)' \neq f'+g'\end{align*}

#### Example A

Find the derivative of:

f(x)=3x2+2x\begin{align*}f(x)=3x^2+2x\end{align*}

Solution

Use the power rule to help:

ddx[3x2+2x]\begin{align*}\frac {d}{dx}\left [{3x^2+2x} \right ]\end{align*} =ddx[3x2]+ddx[2x]\begin{align*}= \frac {d}{dx}\left [{3x^2} \right ]+\frac {d}{dx}\left [{2x} \right ]\end{align*}
=3ddx[x2]+2ddx[x]\begin{align*}= 3\frac {d}{dx}\left [{x^2} \right ]+2 \frac {d}{dx}\left [{x} \right ]\end{align*}
=3[2x]+2[1]\begin{align*}= 3 \left [{2x} \right ]+2 \left [{1} \right ]\end{align*}
=6x+2\begin{align*}= 6x+2\end{align*}

#### Example B

Find the derivative: f(x)=x35x2\begin{align*}f(x)=x^3-5x^2\end{align*}

Solution

Again use the power rule to help:

ddx[x35x2]\begin{align*}\frac {d}{dx}\left [{x^3-5x^2} \right ]\end{align*} =ddx[x3]5ddx[x2]\begin{align*}= \frac {d}{dx}\left [{x^3} \right ]-5 \frac {d}{dx}\left [{x^2} \right ]\end{align*}
=3x25[2x]\begin{align*}= 3x^2- 5 \left [{2x} \right ]\end{align*}
=3x210x\begin{align*}= 3x^2-10x\end{align*}

#### Example C

Find dydx\begin{align*}\frac {dy}{dx}\end{align*} for y=(3x4+2)(7x31)\begin{align*}y = (3x^4 + 2)(7x^3 - 1)\end{align*}

Solution

There two methods to solve this problem. One is to multiply to find 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 = (3x4+2)(7x31)\begin{align*}(3x^4 + 2)(7x^3 -1)\end{align*}
= 21x73x4+14x32\begin{align*}21x^7 -3x^4 + 14x^3 -2\end{align*}
Taking the derivative of the sum yields
dydx\begin{align*}\frac {dy}{dx}\end{align*} =147x612x3+42x2+0\begin{align*}= {147x^6-12x^3+42x^2+0}\end{align*}
=147x612x3+42x2\begin{align*}= 147x^6-12x^3+42x^2\end{align*}
Now we use the product rule.
y\begin{align*}y'\end{align*} =(3x4+2)(7x31)+(3x4+2)(7x31)\begin{align*}= (3x^4+2)\cdot (7x^3-1)'+(3x^4+2)'\cdot (7x^3-1)\end{align*}
=(3x4+2)(21x2)+(12x3)(7x31)\begin{align*}= (3x^4+2)(21x^2)+(12x^3)(7x^3-1)\end{align*}
=(63x6+42x2)+(84x612x3)\begin{align*}= (63x^6+42x^2)+(84x^6-12x^3)\end{align*}
=147x612x3+42x2\begin{align*}= 147x^6-12x^3+42x^2\end{align*}

Concept question wrap-up

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:

a=Fv(δmδt)m\begin{align*}a = \left(\frac{F - v \left(\frac{\delta m}{\delta t}\right)}{m}\right)\end{align*}

Looking at the upper-right portion of the equation, we can see that as mass decreases, the fraction δmδt\begin{align*}\frac{\delta m}{\delta t}\end{align*} goes negative. Since v\begin{align*}-v\end{align*} is multiplied by that fraction, it goes positive, and the overall function increases, meaning the rocket accelerates.

Looks like Juan was right.

### Vocabulary

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.

A differentiable function is one which has a derivative that can be calculated.

### Guided Practice

Questions

Find the Derivative

1) Given: t(x)=x1\begin{align*} t(x) = x - 1\end{align*}

What is dtdx\begin{align*}\frac{dt}{dx}\end{align*} when x=0\begin{align*}x = 0\end{align*}

2) What is the derivative of g(x)=(x1)(x+1)\begin{align*}g(x) = (-x -1)(x + 1)\end{align*}?

3) Given a(x)=πx0.54+6x4\begin{align*} a(x) = -\pi x^{-0.54} + 6x^4\end{align*}

What is dydx\begin{align*}\frac{dy}{dx}\end{align*}

4) Given \begin{align*}y(-2) = 0\end{align*} and \begin{align*}(yc)'(-2) = 132\end{align*}

Find \begin{align*}c (-2)\end{align*} assuming \begin{align*}y'(-2) = 11\end{align*}.

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*} ..... By the power rule
\begin{align*}1' = 0\end{align*} ..... 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*}
By the difference rule: \begin{align*}(-x^2 -2x -1)' = (-x^2)' -(2x)' -(1)' = -2x -2\end{align*}

3) We'll use the difference and power rules:

\begin{align*}\frac{d}{dx}(-\pi x^{-0.54} + 6x^4) =\end{align*}
\begin{align*}\frac{d}{dx}(-\pi x^{-0.54}) + \frac{d}{dx}(6x^4)\end{align*} ..... By the difference rule
\begin{align*}\to 0.54 \pi x^{-1.54} + 24x^3\end{align*} ..... By the power rule

4) We'll apply the product rule:

\begin{align*}(yc)' = y'c + yc'\end{align*}
\begin{align*}(yc)'(-2) = y'(-2)c(-2) + y(-2)c'(-2)\end{align*} ..... By the product rule
\begin{align*}132 = 11c(-2) + (0)c'(-2)\end{align*} ..... Substitute the given values
\begin{align*}132 = 11c(-2)\end{align*} ..... Because \begin{align*}(0)c'(-2) = 0\end{align*}
\begin{align*}12 = c(-2)\end{align*} ..... Simplify

5) We'll use the product rule:

\begin{align*}(pq)' = p'q + pq'\end{align*}.
\begin{align*}p(x) = -5x \to p'(x) = -5\end{align*} .... By the power rule
\begin{align*}q(x) = cos (x) \to q'(x) = -sin (x)\end{align*} ..... By the power rule and simplifying
So we get \begin{align*}[(-5x)cos (x)]' = (-5)cos (x) + (-5x)[-sin (x)]\end{align*}
\begin{align*}= -5cos (x) + (5x)sin (x)\end{align*}

### Practice

Find the Derivative using the sum/difference rule

1. \begin{align*}y = \frac{1} {2} (x^3 - 2x^2 + 1)\end{align*}
2. \begin{align*}y = \sqrt{2} x^3 -\frac{1} {\sqrt{2}}x^2 + 2x + \sqrt{2}\end{align*}
3. \begin{align*}y = a^2 - b^2 + x^2 - a - b + x\end{align*} (where a, b are constants)
4. \begin{align*}y = x^{-3} + \frac{1} {x^7}\end{align*}
5. \begin{align*}y = \sqrt{x} + \frac{1} {\sqrt{x}} \end{align*}
6. \begin{align*}f(x) = (-3x + 4)^2\end{align*}
7. \begin{align*}f(x) = -0.93x^{10} + (\pi^3x)^{\frac{-5}{12}}\end{align*}
8. What is \begin{align*}\frac{d}{dx} (2x + 1)^2\end{align*}?
9. Given: \begin{align*}a(x) = (-5x + 3)^2\end{align*} What is \begin{align*}\frac{dy}{dx}\end{align*}?
10. \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

1. \begin{align*}y = (x^3 - 3x^2 + x) \cdot (2x^3 + 7x^4)\end{align*}
2. \begin{align*}y = \left (\frac{1} {x} + \frac{1} {x^2}\right ) (3x^4 - 7) \end{align*}
3. What is \begin{align*}[(-3x^2 + x + 4)(-3x - 3)]\end{align*}?
4. \begin{align*}v(x) = (3x - 3) \cdot cos (x)\end{align*}
5. 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*}
6. Given \begin{align*}g(x) = (4x^2 - 4x - 5)(3x - 3)\end{align*} Find \begin{align*}g'(2)\end{align*}
7. Find \begin{align*}\frac{d}{dx}[(-4x + 3) \cdot sin (x)\end{align*}
8. Find \begin{align*}\frac{d}{dx}[(x^2 - 3) (-2x^2 + 4x - 1)]\end{align*}
9. 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*}
10. Given \begin{align*}d(x) =(2x^2 + 3x - 1)(2x + 1)\end{align*} Find \begin{align*}d'(-1)\end{align*}

### Vocabulary Language: English

derivative

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 $f'(x)$, $\frac{dy}{dx}$, $y'$, $\frac{df}{dx}$ and \frac{df(x)}{dx}.
differentiable

differentiable

A differentiable function is a function that has a derivative that can be calculated.
product rule

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

theorem

A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.