Find the derivative of each function. f ( t ) = 2 t ( t − 3 + t 2 / 3 ) f(t)=2t(t^{-3}+t^{2/3})

− 4 t − 3 + 10 3 t 2 / 3 -4t^{-3}+\frac{10}{3}t^{2/3} − 4 t − 3 + 3 10 t 2/3

2 t − 3 + 2 t 2 / 3 2t^{-3}+2t^{2/3} 2 t − 3 + 2 t 2/3

− 3 t − 4 + 2 3 t − 1 / 3 -3t^{-4}+\frac23t^{-1/3} − 3 t − 4 + 3 2 t − 1/3

− 6 t − 3 + 4 3 t 2 / 3 -6t^{-3}+\frac43t^{2/3}

Find the derivative of the function. f ( x ) = 2 x − 1 x 3 + 2 f\left(x\right)=\frac{2x-1}{x^3+2}

− 4 x 3 + 3 x 2 + 4 ( x 3 + 2 ) 2 \frac{-4x^3+3x^2+4}{(x^3+2)^2}

4 x 3 − 3 x 2 − 4 ( x 3 + 2 ) 2 \frac{4x^3-3x^2-4}{(x^3+2)^2}

4 x 3 − 3 x 2 − 4 ( 2 x − 1 ) 2 \frac{4x^3-3x^2-4}{(2x-1)^2}

8 x 3 + 3 x 2 + 4 ( x 3 + 2 ) 2 \frac{8x^3+3x^2+4}{(x^3+2)^2}

Find the derivative of the function. g ( y ) = 5 y 2 + 2 y − 1 y 2 − 2 g(y)=\frac{5y^2+2y-1}{y^2-2}

− 2 y 2 − 18 y − 4 ( y 2 − 2 ) 2 \frac{-2y^2-18y-4}{(y^2-2)^2} ( y 2 − 2 ) 2 − 2 y 2 − 18 y − 4

2 y 2 + 18 y + 4 ( 5 y 2 + 2 y − 1 ) 2 \frac{2y^2+18y+4}{(5y^2+2y-1)^2} ( 5 y 2 + 2 y − 1 ) 2 2 y 2 + 18 y + 4

− 2 y 2 − 18 y ( y 2 − 2 ) 2 \frac{-2y^2-18y}{(y^2-2)^2} ( y 2 − 2 ) 2 − 2 y 2 − 18 y

3. Techniques of Differentiation

Product and Quotient Rules

3. Techniques of Differentiation

Product and Quotient Rules: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

To find the derivative of functions multiplied or divided, use the product rule and quotient rule, respectively. The product rule states that for functions f(x) and g(x), the derivative is $f' (x) g (x) + f (x) g' (x)$ . The quotient rule is $g (x) f' (x) - f (x) g' (x)$ over $g (x)^2$ . Mastering these rules is essential for calculus proficiency.

concept

The Product Rule

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The Product Rule Video Summary

When dealing with the derivatives of functions, it's essential to understand how to apply the product rule, especially when two functions are multiplied together. The product rule states that if you have two functions, f(x) and g(x), their derivative can be found using the formula:

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

This means you take the first function f(x) and multiply it by the derivative of the second function g'(x), and then add it to the second function g(x) multiplied by the derivative of the first function f'(x).

A helpful mnemonic to remember this is: "left d right plus right d left," where "d" indicates taking the derivative. This approach allows for a systematic way to find the derivative without needing to expand the functions first.

For example, consider the function h(x) = (x - 5)(2x + 9). To find the derivative h'(x), apply the product rule:

1. Identify f(x) = x - 5 and g(x) = 2x + 9.

2. Calculate g'(x) = 2 and f'(x) = 1.

3. Apply the product rule:

h'(x) = (x - 5)(2) + (2x + 9)(1)

4. Simplify the expression:

h'(x) = 2x - 10 + 2x + 9 = 4x - 1

Now, let's look at another example with the function y = (2x² - 1)(3 + x³). Using the product rule:

1. Identify f(x) = 2x² - 1 and g(x) = 3 + x³.

2. Calculate g'(x) = 3x² and f'(x) = 4x.

3. Apply the product rule:

y' = (2x² - 1)(3x²) + (3 + x³)(4x)

4. Distribute and simplify:

y' = 6x⁴ - 3x² + 12x + 4x⁴ = 10x⁴ - 3x² + 12x

Understanding and practicing the product rule is crucial for mastering calculus, as it provides a reliable method for differentiating products of functions efficiently.

example

The Product Rule Example 1

Video duration:

The Product Rule Example 1 Video Summary

To find the derivative of the function $ y = 4x^2(8 - x^3) $, we can use two methods: the product rule and the power rule after expanding the expression.

First, let's apply the product rule. The product rule states that if you have two functions $ u $ and $ v $, the derivative $ y' $ is given by:

\[ y' = u'v + uv' \]

In our case, let $ u = 4x^2 $ and $ v = 8 - x^3 $. We need to find the derivatives $ u' $ and $ v' $.

Calculating $ u' $:

\[ u' = \frac{d}{dx}(4x^2) = 8x \]

Calculating $ v' $:

\[ v' = \frac{d}{dx}(8 - x^3) = -3x^2 \]

Now, substituting back into the product rule formula:

\[ y' = (4x^2)(-3x^2) + (8 - x^3)(8x) \]

Calculating the first term:

\[ -12x^4 \]

Calculating the second term:

\[ 64x - 8x^4 \]

Combining these results gives:

\[ y' = -12x^4 + 64x - 8x^4 = -20x^4 + 64x \]

Now, let's use the second method by expanding the expression first. Distributing $ 4x^2 $ into $ (8 - x^3) $ gives:

\[ y = 32x^2 - 4x^5 \]

Now we can apply the power rule directly. The power rule states that:

\[ \frac{d}{dx}(x^n) = nx^{n-1} \]

Calculating the derivative:

For $ 32x^2 $:

\[ \frac{d}{dx}(32x^2) = 64x \]

For $ -4x^5 $:

\[ \frac{d}{dx}(-4x^5) = -20x^4 \]

Combining these results gives:

\[ y' = 64x - 20x^4 \]

Both methods yield the same derivative:

\[ y' = -20x^4 + 64x \]

This demonstrates that while the product rule is often quicker, expanding the expression and using the power rule is a valid alternative for verifying results.

Problem

Find the derivative of each function.
$y=(3x+5)^2$

$3$

$6x$

$9x+15$

$18x+30$

Problem

Find the derivative of each function.
$f (t) = 2 t (t^{- 3} + t^{2 / 3})$

$2t^{-3}+2t^{2/3}$

$-3t^{-4}+\frac23t^{-1/3}$

$-4t^{-3}+\frac{10}{3}t^{2/3}$

$- 6 t^{- 3} + \frac{4}{3} t^{2 / 3}$

example

The Product Rule Example 2

Video duration:

The Product Rule Example 2 Video Summary

To find the slope of the tangent line to the curve defined by the function $ y = x^2 - 5(6x + 1) $ at the point where $ x = 2 $, we first need to determine the derivative of the function. The slope of the tangent line at any point on a curve is given by the derivative of the function at that point.

In this case, the function can be expressed as a product of two functions: $ u = x^2 - 5 $ and $ v = 6x + 1 $. To find the derivative $ y' $, we apply the product rule, which states that if $ y = uv $, then $ y' = u'v + uv' $.

Calculating the derivatives of $ u $ and $ v $:

The derivative of $ u $ is $ u' = 2x $ (using the power rule).
The derivative of $ v $ is $ v' = 6 $ (since the derivative of $ 6x + 1 $ is simply 6).

Now, applying the product rule:

$ y' = (x^2 - 5)(6) + (6x + 1)(2x) $

Distributing the terms gives:

$ y' = 6(x^2 - 5) + (6x + 1)(2x) $

$ = 6x^2 - 30 + 12x^2 + 2x $

Combining like terms results in:

$ y' = 18x^2 + 2x - 30 $

Next, we need to evaluate the derivative at $ x = 2 $ to find the slope of the tangent line:

$ y'(2) = 18(2^2) + 2(2) - 30 $

$ = 18(4) + 4 - 30 $

$ = 72 + 4 - 30 $

$ = 76 - 30 $

$ = 46 $

Thus, the slope of the tangent line to the curve at $ x = 2 $ is $ 46 $. This process illustrates the application of the product rule and the evaluation of derivatives to find the slope at a specific point on a curve.

concept

The Quotient Rule

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The Quotient Rule Video Summary

When dealing with the differentiation of functions, it's essential to understand that the rules for multiplication and division differ significantly. For two functions being divided, we utilize the quotient rule. This rule is crucial for finding the derivative of a function expressed as the ratio of two other functions.

To apply the quotient rule, consider a function h(x) = \frac{f(x)}{g(x)}, where f(x) is the numerator and g(x) is the denominator. The quotient rule states that the derivative of this function can be calculated using the formula:

\[h'(x) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2}\]

To help remember this formula, a useful mnemonic is: "low d high minus high d low over the square of what's below." Here, "low" refers to the denominator g(x), "high" refers to the numerator f(x), and "d" indicates taking the derivative.

For example, if we have h(x) = \frac{x}{3x - 4}, we identify f(x) = x and g(x) = 3x - 4. Applying the quotient rule:

\[h'(x) = \frac{(3x - 4) \cdot 1 - x \cdot 3}{(3x - 4)^2}\]

After simplifying, we find:

\[h'(x) = \frac{-4}{(3x - 4)^2}\]

In another example, consider y = \frac{2x^2 - 1}{3 - x^3}. Here, we again apply the quotient rule:

\[y' = \frac{(3 - x^3) \cdot (4x) - (2x^2 - 1) \cdot (-3x^2)}{(3 - x^3)^2}\]

After performing the necessary algebraic simplifications, we arrive at:

\[y' = \frac{-2x^4 - 3x^2 + 12x}{(3 - x^3)^2}\]

It's important to note that while simplifying, the denominator can often be left in its squared form without expanding, unless there are clear opportunities for further simplification. Mastery of the quotient rule, along with consistent practice, will enhance your ability to differentiate complex functions effectively.

Problem

Find the derivative of the function.
$y = \frac{2 - 3 t}{4 t^{2} + 7}$

$\frac{12 t^{2} - 16 t - 21}{(4 t^{2} + 7)^{2}}$

$\frac{- 12 t^{2} + 16 t + 21}{(4 t^{2} + 7)^{2}}$

$\frac{- 12 t^{2} + 16 t + 21}{(2 - 3 t)^{2}}$

$\frac{12 t^{2} - 16 t - 21}{(2 - 3 t)^{2}}$

Problem

Find the derivative of the function.
$f (x) = \frac{2 x - 1}{x^{3} + 2}$

$\frac{4 x^{3} - 3 x^{2} - 4}{(x^{3} + 2)^{2}}$

$\frac{4 x^{3} - 3 x^{2} - 4}{(2 x - 1)^{2}}$

$\frac{- 4 x^{3} + 3 x^{2} + 4}{(x^{3} + 2)^{2}}$

$\frac{8 x^{3} + 3 x^{2} + 4}{(x^{3} + 2)^{2}}$

Problem

Find the derivative of the function.
$g (y) = \frac{5 y^{2} + 2 y - 1}{y^{2} - 2}$

$\frac{2 y^{2} + 18 y + 4}{(y^{2} - 2)^{2}}$

$\frac{-2y^2-18y-4}{(y^2-2)^2}$

$\frac{2y^2+18y+4}{(5y^2+2y-1)^2}$

$\frac{-2y^2-18y}{(y^2-2)^2}$

example

The Quotient Rule Example 3

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The Quotient Rule Example 3 Video Summary

To find the derivative of the function $ y = \frac{(x + 1)(x^2 - 3x)}{x^3} $, we will apply the quotient rule, which states that if you have a function in the form of $ \frac{u}{v} $, the derivative is given by:

\[\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\]

In this case, let $ u = (x + 1)(x^2 - 3x) $ and $ v = x^3 $. We first need to find the derivatives of $ u $ and $ v $.

The derivative of $ v $ is straightforward:

\[\frac{dv}{dx} = 3x^2\]

Next, we need to find $ \frac{du}{dx} $. Since $ u $ is a product of two functions, we will use the product rule:

\[\frac{du}{dx} = (x + 1) \frac{d}{dx}(x^2 - 3x) + (x^2 - 3x) \frac{d}{dx}(x + 1)\]

Calculating the derivatives:

\[\frac{d}{dx}(x^2 - 3x) = 2x - 3\]

\[\frac{d}{dx}(x + 1) = 1\]

Substituting these back into the product rule gives:

\[\frac{du}{dx} = (x + 1)(2x - 3) + (x^2 - 3x)(1)\]

Now, we can expand this expression:

\[= (x + 1)(2x - 3) + (x^2 - 3x)\]

Expanding $ (x + 1)(2x - 3) $ using the distributive property:

\[= 2x^2 - 3x + 2x - 3 + x^2 - 3x\]

Combining like terms results in:

\[= 3x^2 - 4x - 3\]

Now we have $ \frac{du}{dx} = 3x^2 - 4x - 3 $ and $ \frac{dv}{dx} = 3x^2 $. We can substitute these into the quotient rule formula:

\[\frac{dy}{dx} = \frac{x^3(3x^2 - 4x - 3) - (x + 1)(x^2 - 3x)(3x^2)}{(x^3)^2}\]

Next, we simplify the numerator. First, distribute $ x^3 $ into $ (3x^2 - 4x - 3) $:

\[= 3x^5 - 4x^4 - 3x^3\]

Now, we need to expand $ (x + 1)(x^2 - 3x)(3x^2) $:

\[= 3x^2(x + 1)(x^2 - 3x)\]

Expanding $ (x + 1)(x^2 - 3x) $ gives:

\[= x^3 - 3x^2 + x^2 - 3x = x^3 - 2x^2 - 3x\]

Now multiplying by $ 3x^2 $:

\[= 3x^2(x^3 - 2x^2 - 3x) = 3x^5 - 6x^4 - 9x^3\]

Substituting this back into the numerator gives:

\[3x^5 - 4x^4 - 3x^3 - (3x^5 - 6x^4 - 9x^3)\]

Combining like terms results in:

\[= 2x^4 + 6x^3\]

Thus, the derivative simplifies to:

\[\frac{dy}{dx} = \frac{2x^4 + 6x^3}{x^6}\]

Finally, we can simplify this expression:

\[= \frac{2x^4}{x^6} + \frac{6x^3}{x^6} = \frac{2}{x^2} + \frac{6}{x^3}\]

Therefore, the final result for the derivative is:

\[\frac{dy}{dx} = \frac{2}{x^2} + \frac{6}{x^3}\]

example

The Quotient Rule Example 4

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The Quotient Rule Example 4 Video Summary

In this problem, we are tasked with finding the derivative of a function $ h(x) $ defined in two different ways, using the information provided about two other functions, $ f(x) $ and $ g(x) $. We know that $ f(2) = 4 $, $ f'(2) = -1 $, $ g(2) = 3 $, and $ g'(2) = 0 $. We will apply the product rule and the quotient rule to find $ h'(2) $ for each scenario.

For the first scenario, where $ h(x) = f(x) \cdot g(x) $, we utilize the product rule, which states that the derivative of a product of two functions is given by:

\[h'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x)\]

To find $ h'(2) $, we substitute $ x = 2 $ into the derivative:

\[h'(2) = f'(2) \cdot g(2) + f(2) \cdot g'(2)\]

Substituting the known values:

\[h'(2) = (-1) \cdot 3 + 4 \cdot 0\]

This simplifies to:

\[h'(2) = -3 + 0 = -3\]

Thus, for the first scenario, $ h'(2) = -3 $.

In the second scenario, where $ h(x) = \frac{f(x)}{g(x)} $, we apply the quotient rule, which states that the derivative of a quotient of two functions is given by:

\[h'(x) = \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{(g(x))^2}\]

To find $ h'(2) $, we substitute $ x = 2 $ into the derivative:

\[h'(2) = \frac{g(2) \cdot f'(2) - f(2) \cdot g'(2)}{(g(2))^2}\]

Substituting the known values:

\[h'(2) = \frac{3 \cdot (-1) - 4 \cdot 0}{(3)^2}\]

This simplifies to:

\[h'(2) = \frac{-3 - 0}{9} = \frac{-3}{9} = -\frac{1}{3}\]

Thus, for the second scenario, $ h'(2) = -\frac{1}{3} $.

In summary, we found that when $ h(x) = f(x) \cdot g(x) $, $ h'(2) = -3 $, and when $ h(x) = \frac{f(x)}{g(x)} $, $ h'(2) = -\frac{1}{3} $. This problem illustrates the application of the product and quotient rules in calculus, demonstrating how to derive the rates of change of composite functions using known values of their components.

example

The Quotient Rule Example 5

Video duration:

The Quotient Rule Example 5 Video Summary

In this problem, we analyze the population of a bird species represented by the function $ p(t) = \frac{6t + 5}{0.3t^2 + 2} $, where $ t $ denotes time in years. To understand how the population changes over time, we need to find the derivative $ p'(t) $, which indicates the rate of change of the population with respect to time.

To compute the derivative, we apply the quotient rule, which states that if you have a function in the form of $ \frac{f(t)}{g(t)} $, the derivative is given by:

\[ p'(t) = \frac{g(t)f'(t) - f(t)g'(t)}{(g(t))^2} \]

Here, $ f(t) = 6t + 5 $ and $ g(t) = 0.3t^2 + 2 $. The derivatives are $ f'(t) = 6 $ and $ g'(t) = 0.6t $. Applying the quotient rule, we have:

\[ p'(t) = \frac{(0.3t^2 + 2)(6) - (6t + 5)(0.6t)}{(0.3t^2 + 2)^2} \]

Next, we simplify the numerator:

1. The first term: $ 6(0.3t^2 + 2) = 1.8t^2 + 12 $

2. The second term: $ (6t + 5)(0.6t) = 3.6t^2 + 3t $

Combining these, we get:

\[ p'(t) = \frac{(1.8t^2 + 12) - (3.6t^2 + 3t)}{(0.3t^2 + 2)^2} \]

Distributing the negative sign gives:

\[ p'(t) = \frac{-1.8t^2 - 3t + 12}{(0.3t^2 + 2)^2} \]

This expression represents the derivative $ p'(t) $, which indicates how the population of the bird species changes over time. Specifically, $ p'(t) $ provides the rate of change of the population at any given time $ t $. For instance, evaluating $ p'(2) $ would yield the rate of change of the bird population at 2 years, allowing us to understand whether the population is increasing or decreasing at that specific time.

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Product and Quotient Rules

3. Techniques of Differentiation

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3. Techniques of Differentiation

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Here’s what students ask on this topic:

What is the product rule in calculus?

The product rule in calculus is a formula used to find the derivative of the product of two functions. If you have two functions, f(x) and g(x), the product rule states that the derivative of their product is given by:

$(f g)' = f' g + f g'$

This means you take the derivative of the first function, multiply it by the second function as is, and then add the first function as is multiplied by the derivative of the second function.

How do you apply the quotient rule in calculus?

The quotient rule in calculus is used to find the derivative of a function that is the quotient of two other functions. If you have two functions, f(x) and g(x), where g(x) ≠ 0, the quotient rule states that the derivative of their quotient is given by:

$(\frac{f}{g})' = \frac{g f' - f g'}{g^2}$

This means you take the derivative of the numerator, multiply it by the denominator as is, subtract the numerator as is multiplied by the derivative of the denominator, and then divide the entire expression by the square of the denominator.

Can you provide an example of using the product rule?

Sure! Let's find the derivative of the function h(x) = (x - 5)(2x + 9) using the product rule. Here, f(x) = x - 5 and g(x) = 2x + 9. According to the product rule:

$(f g)' = f' g + f g'$

First, find the derivatives: f'(x) = 1 and g'(x) = 2. Then apply the product rule:

$(x - 5)' (2 x + 9) + (x - 5) (2)$

This simplifies to:

$1 (2 x + 9) + (x - 5) 2$

Which further simplifies to:

$2 x + 9 + 2 x - 10$

Combining like terms, we get:

$4 x - 1$

So, the derivative h'(x) = 4x - 1.

What is the mnemonic device for remembering the quotient rule?

A helpful mnemonic device for remembering the quotient rule is: "Low d high minus high d low over the square of what's below." This means:

$(\frac{f}{g})' = \frac{g f' - f g'}{g^2}$

Where "low" refers to the denominator function g(x), "high" refers to the numerator function f(x), and "d" means taking the derivative. This mnemonic helps you remember the order and operations needed to apply the quotient rule correctly.

Can you provide an example of using the quotient rule?

Sure! Let's find the derivative of the function h(x) = x / (3x - 4) using the quotient rule. Here, f(x) = x and g(x) = 3x - 4. According to the quotient rule:

$(\frac{f}{g})' = \frac{g f' - f g'}{g^2}$

First, find the derivatives: f'(x) = 1 and g'(x) = 3. Then apply the quotient rule:

$(\frac{x}{3 x - 4})' = \frac{(3 x - 4) 1 - x 3}{(3 x - 4)^2}$

This simplifies to:

$\frac{3 x - 4 - 3 x}{(3 x - 4)^2}$

Combining like terms, we get:

$\frac{- 4}{(3 x - 4)^2}$

So, the derivative h'(x) = -4 / (3x - 4)^2.

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Additional resources for Product and Quotient Rules

PRACTICE PROBLEMS AND ACTIVITIES (83)

Product and Quotient Rules: Videos & Practice Problems

The Product Rule

The Product Rule Video Summary

The Product Rule Example 1

The Product Rule Example 1 Video Summary

Find the derivative of each function.y=(3x+5)2y=(3x+5)^2y=(3x+5)2

Find the derivative of each function.f(t)=2t(t−3+t2/3)f(t)=2t(t^{-3}+t^{2/3})

The Product Rule Example 2

The Product Rule Example 2 Video Summary

The Quotient Rule

The Quotient Rule Video Summary

Find the derivative of the function.y=2−3t4t2+7y=\frac{2-3t}{4t^2+7}

Find the derivative of the function.f(x)=2x−1x3+2f\left(x\right)=\frac{2x-1}{x^3+2}

Find the derivative of the function.g(y)=5y2+2y−1y2−2g(y)=\frac{5y^2+2y-1}{y^2-2}

The Quotient Rule Example 3

The Quotient Rule Example 3 Video Summary

The Quotient Rule Example 4

The Quotient Rule Example 4 Video Summary

The Quotient Rule Example 5

The Quotient Rule Example 5 Video Summary

Do you want more practice?

Here’s what students ask on this topic:

What is the product rule in calculus?

How do you apply the quotient rule in calculus?

Can you provide an example of using the product rule?

What is the mnemonic device for remembering the quotient rule?

Can you provide an example of using the quotient rule?

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Find the derivative of each function.
$y=(3x+5)^2$

Find the derivative of each function.
$f (t) = 2 t (t^{- 3} + t^{2 / 3})$

Find the derivative of the function.
$y = \frac{2 - 3 t}{4 t^{2} + 7}$

Find the derivative of the function.
$f (x) = \frac{2 x - 1}{x^{3} + 2}$

Find the derivative of the function.
$g (y) = \frac{5 y^{2} + 2 y - 1}{y^{2} - 2}$