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8.10: Derivatives of Sums and Differences

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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?

Watch This

Embedded Video:

- Khan Academy: Product Rule

Guidance

Derivatives of Sums and Differences

Theorem: If f and g are two differentiable functions at x then

\frac {d}{dx}\left [{f(x)+g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]+\frac {d}{dx}\left [{g(x)} \right ]

and

\frac {d}{dx}\left [{f(x)-g(x)} \right ]=\frac{d}{dx}\left [{f(x)} \right ]- \frac {d}{dx}\left [{g(x)} \right ]

In simpler notation

(f \pm g)'= f'\pm g' .

The Product Rule

Theorem: (The Product Rule) If f and g are differentiable at x , then

\frac {d}{dx}\left [{f(x)\cdot g(x)} \right ]= f(x) \frac {d}{dx}g(x) + g(x)\frac {d}{dx}f(x)

In a simpler notation

(f\cdot g)'=f\cdot g'+g\cdot f'

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 (f\cdot g)' \neq f'+g'

Example A

Find the derivative of:

f(x)=3x^2+2x

Solution

Use the power rule to help:

\frac {d}{dx}\left [{3x^2+2x} \right ] = \frac {d}{dx}\left [{3x^2} \right ]+\frac {d}{dx}\left [{2x} \right ]
= 3\frac {d}{dx}\left [{x^2} \right ]+2 \frac {d}{dx}\left [{x} \right ]
= 3 \left [{2x} \right ]+2 \left [{1} \right ]
= 6x+2

Example B

Find the derivative: f(x)=x^3-5x^2

Solution

Again use the power rule to help:

\frac {d}{dx}\left [{x^3-5x^2} \right ] = \frac {d}{dx}\left [{x^3} \right ]-5 \frac {d}{dx}\left [{x^2} \right ]
= 3x^2- 5 \left [{2x} \right ]
= 3x^2-10x

Example C

Find \frac {dy}{dx} for y = (3x^4 + 2)(7x^3 - 1)

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 = (3x^4 + 2)(7x^3 -1)
= 21x^7 -3x^4 + 14x^3 -2
Taking the derivative of the sum yields
\frac {dy}{dx} = {147x^6-12x^3+42^2+0}
= 147x^6-12x^3+42x^2
Now we use the product rule.
y' = (3x^4+2)\cdot (7x^3-1)'+(3x^4+2)'\cdot (7x^3-1)
= (3x^4+2)(21x^2)+(12x^3)(7x^3-1)
= (63x^6+42x^2)+(84x^6-12x^3)
= 147x^6-12x^3+42x^2
Which is the same answer.

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 = \left(\frac{F - v \left(\frac{\delta m}{\delta t}\right)}{m}\right)

Looking at the upper-right portion of the equation, we can see that as mass decreases, the fraction \frac{\delta m}{\delta t} goes negative. Since -v 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 \cdot the derivative of the second function, added to the second function \cdot 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) = x - 1

What is \frac{dt}{dx} when x = 0

2) What is the derivative of g(x) = (-x -1)(x + 1) ?

3) Given  a(x) = -\pi x^{-0.54} + 6x^4

What is \frac{dy}{dx}

4) Given y(-2) = 0 and (yc)'(-2) = 132

Find c (-2) assuming y'(-2) = 11 .

5) What is \frac{d}{dx}[(-5x)(cos x)]?

Solutions

1) By the difference rule: (x - 1)' = (x)' - (1)' = 0

x' = 1 ..... By the power rule
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 (-x -1)(x + 1) \to -x^2 -2x -1.
By the difference rule: (-x^2 -2x -1)' = (-x^2)' -(2x)' -(1)' = -2x -2

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

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

4) We'll apply the product rule:

(yc)' = y'c + yc'
(yc)'(-2) = y'(-2)c(-2) + y(-2)c'(-2) ..... By the product rule
132 = 11c(-2) + (0)c'(-2) ..... Substitute the given values
132 = 11c(-2) ..... Because (0)c'(-2) = 0
12 = c(-2) ..... Simplify

5) We'll use the product rule:

(pq)' = p'q + pq' .
p(x) = -5x \to p'(x) = -5 .... By the power rule
q(x) = cos x \to q'(x) = -sin x ..... By the power rule and simplifying
So we get [(-5x)(cos x)]' = (-5)(cos x) + (-5x)(-sin x)
= -5cos x + 5xsin x

Practice

Find the Derivative using the sum/difference rule

  1. y = \frac{1} {2} (x^3 - 2x^2 + 1)
  2. y = \sqrt{2} x^3 -\frac{1} {\sqrt{2}}x^2 + 2x + \sqrt{2}
  3. y = a^2 - b^2 + x^2 - a - b + x (where a , b are constants)
  4. y = x^{-3} + \frac{1} {x^7}
  5. y = \sqrt{x} + \frac{1} {\sqrt{x}}
  6. f(x) = (-3x + 4)^2
  7. f(x) = -0.93x^{10} + (\pi^3x)^{\frac{-5}{12}}
  8. What is \frac{d}{dx} (2x + 1)^2
  9. Given: a(x) = (-5x + 3)^2 What is \frac{dy}{dx}
  10.  v(x) = -3x^3 + 5x^2 - 2x - 3 What is v'(0)

Find the Derivative using the product rule

  1. y = (x^3 - 3x^2 + x) \cdot (2x^3 + 7x^4)
  2. y = \left (\frac{1} {x} + \frac{1} {x^2}\right ) (3x^4 - 7)
  3. What is [(-3x^2 + x + 4)(-3x - 3)]
  4. v(x) = (3x - 3)(cos x)
  5. Given: k(-2) = 0 k'(-2) = 18 Find r(-2) when (kr)'(-2) = 54
  6. Given g(x) = (4x^2 - 4x - 5)(3x - 3) Find g'(2)
  7. Find \frac{d}{dx}[(-4x + 3)(sin x)
  8. Find \frac{d}{dx}[(x^2 - 3) (-2x^2 + 4x - 1)]
  9. Given t(1) = 0 t'(1) = 17 Find a(1) when (ta)'(1) = 272
  10. Given d(x) =(2x^2 + 3x - 1)(2x + 1) Find d'(-1)

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Date Created:

Nov 01, 2012

Last Modified:

Aug 04, 2014

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