The Product Rule

Section 3.1 The Product Rule

The rule for differentiating a product of functions is slightly more complicated than the previous rules.

Example 3.1.1. The Derivative of a Product Is Not the Product of the Derivatives.

One might at first glance think that the derivative of a product of two functions should be the product of the derivatives. Especially because the derivative of a sum is the sum of the derivatives (sum rule). Unfortunately, this is not true. In other words,

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\begin{equation*} \frac{d}{dx} \brac{f(x) g(x)} \neq f'(x) \cdot g'(x) \end{equation*}

For example, if $f(x)=x^2$ and $g(x)=x^3\text{,}$ the derivative of the product is,

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\begin{align*} \frac{d}{dx} \brac{f(x) g(x)} \amp= \frac{d}{dx} \brac{x^2 \cdot x^3}\\ \amp= \frac{d}{dx} x^5\\ \amp= 5x^4 \end{align*}

However, $f'(x) = 2x$ and $g'(x) = 3x^2\text{,}$ so $f'(x) \cdot g'(x) = 6x^3\text{,}$ which is not the same as $5x^4\text{.}$

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Theorem 3.1.2. Product Rule.

\begin{equation*} \boxed{(fg)' = f'g + fg'} \quad \text{"$fg$ prime equals $f$ prime times $g$ plus $f$ times $g$ prime"} \end{equation*}

Or, with the function arguments,

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\begin{equation*} \brac{f(x) \cdot g(x)}' = f'(x)g(x) + f(x)g'(x) \end{equation*}

Or,

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\begin{equation*} \frac{d}{dx} \brac{f(x) g(x)} = \brac{\frac{d}{dx}f(x)} \cdot g(x) + f(x) \cdot \brac{\frac{d}{dx}g(x)} \end{equation*}

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Intuitively, if we call $f$ the 1st function and $g$ the 2nd function, then it says the derivative of their product is,

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\begin{equation*} \boxed{\text{"Derivative of the 1st"} \times \text{"2nd"} + \text{"1st"} \times \text{"Derivative of the 2nd"}} \end{equation*}

It’s like writing the product twice, and each time you take the derivative of exactly one of the functions.

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Remark 3.1.3.

Note that because the two terms are added, their order doesn’t matter. So $(fg)' = f'g + fg' = fg' + f'g\text{.}$

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Example 3.1.4. Simple Product (Expanding Also Works).

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

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Solution.

This is the product of $2x+1$ and $x^2-3\text{,}$ so the product rule says,

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\begin{align*} f'(x) \amp= \underbrace{2}_{\text{deriv of 1st}} \cdot \underbrace{(x^2-3)}_{\text{2nd}} + \underbrace{(2x+1)}_{\text{1st}} \cdot \underbrace{2x}_{\text{deriv of 2nd}} \end{align*}

Then, we can expand and simplify,

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\begin{align*} f'(x) \amp= 2x^2 - 6 + 4x^2 + 2x\\ \amp= 6x^2 + 2x - 6 \end{align*}

In this case, you may recognize that we could have expanded the function first, and then differentiated term by term,

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\begin{align*} f(x) \amp= (2x+1)(x^2-3)\\ \amp= 2x^3 + x^2 - 6x - 3 \end{align*}

Then,

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\begin{equation*} f'(x) = 6x^2 + 2x - 6 \end{equation*}

Both methods give the same answer. Here, expanding first is a bit simpler, because the expansion isn’t too challenging.

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Example 3.1.5. More Complicated Product.

Find the derivative of $f(x) = (x^2+3x+1)(x^3-2x+4)\text{.}$

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Solution.

In this case, expanding is more complicated, so the product rule would work better.

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\begin{align*} f'(x) \amp= \underbrace{(2x+3)}_{\text{deriv of 1st}} \cdot (x^3-2x+4) + (x^2+3x+1) \cdot \underbrace{(3x^2-2)}_{\text{deriv of 2nd}} \end{align*}

This is a valid final answer. If we need the derivative at a specific point, say $x = 1\text{,}$ we can plug in directly without expanding,

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\begin{align*} f'(1) \amp= (2+3)(1-2+4) + (1+3+1)(3-2)\\ \amp= 5(3) + 5(1) = 20 \end{align*}

This is much faster than trying to expand everything first.

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Example 3.1.6. Product with Radicals.

Differentiate $f(x) = (x - \sqrt[3]{x})(\sqrt{x}+2)\text{.}$

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Solution.

Expanding is a bit complicated here, because there are 2 radicals. Instead, we can use the product rule,

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\begin{align*} f'(x) \amp= \underbrace{\brac{1 - \frac{1}{3}x^{-2/3}}}_{\text{deriv of 1st}} \cdot (\sqrt{x}+2) + (x - \sqrt[3]{x}) \cdot \underbrace{\frac{1}{2}x^{-1/2}}_{\substack{\text{deriv of} \\ \text{2nd}}} \end{align*}

This is a valid final answer, however, if you wanted to plug in a specific value of $x\text{,}$ it is helpful to rewrite the exponents as positive,

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\begin{align*} f'(x) \amp= \brac{1 - \frac{1}{3x^{2/3}}} \cdot (\sqrt{x}+2) + (x - \sqrt[3]{x}) \cdot \frac{1}{2\sqrt{x}} \end{align*}

Note that $x^{-1/2} = \frac{1}{\sqrt{x}}$ and $x^{-2/3} = \frac{1}{x^{2/3}}\text{.}$

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In general,

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Expand first if it’s simple to do, and would make the function simpler.
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Use the product rule if the expansion is complicated and would take a long time.
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Later on, we’ll see that there are many situations where you can’t expand.

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Subsection 3.1.1 Examples

Exercise Group 3.1.1. Product Rule (No Simplification).

Find each derivative, using the product rule. No need to simplify your answer.

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(a)

$f(x) = (2x-1)(x^2+1)$

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Answer.

$f'(x) = 2(x^2+1)+(2x-1)(2x)$

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(b)

$f(x) = x(3x-8)$

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Answer.

$f'(x) = 1\cdot(3x-8)+x\cdot 3$

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(c)

$f(x) = (x^3+x^2+1)(x^2+2)$

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Answer.

$f'(x) = (3x^2+2x)(x^2+2)+(x^3+x^2+1)(2x)$

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(d)

$f(x) = x^2(1+x-3x^2)$

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Answer.

$f'(x) = 2x(1+x-3x^2)+x^2(1-6x)$

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(e)

$f(x) = (x^4+x^2-1)(x^2-2)$

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Answer.

$f'(x) = (4x^3+2x)(x^2-2)+(x^4+x^2-1)(2x)$

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Exercise Group 3.1.2. Product Rule Two Ways.

Find each derivative in two ways: using the product rule, and by expanding first. Verify that both answers are equal, by simplifying.

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(a)

$f(x) = (x+1)(x^2-x+1)$

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Answer.

$f'(x) = 3x^2$

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(b)

$f(x) = (3x+4)(x-1)$

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Answer.

$f'(x) = 6x + 1$

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(c)

$f(x) = (2x+3)(5x^2-4x)$

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Answer.

$f'(x) = 30x^2+14x-12$

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(d)

$f(x) = x(x-1)$

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Answer.

$f'(x) = 2x - 1$

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(e)

$f(x) = (3-x^2)(x^3-x+1)$

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Answer.

$f'(x) = -5x^4+12x^2-2x-3$

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(f)

$f(x)=(1+2x^2)(x-x^2)$

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Answer.

$f'(x)=1-2x+6x^2-8x^3$

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(g)

$f(x) = (x^3+4x^2+x)(x-1)$

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Answer.

$f'(x) = 4x^3 + 9x^2 - 6x - 1$

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Exercise Group 3.1.3. Evaluate at a Point.

Evaluate each derivative at the given point.

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(a)

$y=(2+7x)(x-3)\text{,}$ $x=2$

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Answer.

$9$

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(b)

$y=(1-2x)(1+2x)\text{,}$ $x=\frac{1}{2}$

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Answer.

$-4$

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(c)

$f(x)=(5x^3+7x^2+3)(2x^2+x+6)\text{,}$ find $f'(-1)$

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Answer.

$f'(-1)=-8$

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(d)

$y=(3-2x-x^2)(x^2+x-2)\text{,}$ $x=-2$

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Answer.

$-9$

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(e)

$y = (1-2x)(3x-4)\text{,}$ $x = 2$

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Answer.

$-13$

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(f)

$y = (1-x+x^2)(x-2)\text{,}$ $x = 1$

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Answer.

$0$

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(g)

$y = x^4(4x^3+2)\text{,}$ $x = -1$

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Answer.

$20$

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Exercise Group 3.1.4. Tangent Lines.

Find the equation of the tangent line to the given curve at the given point.

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(a)

$y=(x^3-5x+2)(3x^2-2x)$ at $(1,-2)$

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Answer.

$y=-10x+8$

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(b)

$y = (2-\sqrt{x})(1+\sqrt{x}+3x)$ at $(1,5)$

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Answer.

$y = x + 4$

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Exercise Group 3.1.5. Differentiate and Simplify.

Find each derivative and simplify.

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(a)

$f(x)=(x^2-3x)(x^5+2)$

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Answer.

$f'(x)=7x^6-18x^5+4x-6$

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(b)

$f(x)=x(x-4)$

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Answer.

$f'(x)=2x-4$

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(c)

$f(x)=x^2(2x-1)$

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Answer.

$f'(x)=6x^2-2x$

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(d)

$f(x)=(3x+2)(2x-7)$

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Answer.

$f'(x)=12x-17$

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(e)

$f(x)=(5x^7+1)(x^2-2x)$

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Answer.

$f'(x)=45x^8-80x^7+2x-2$

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(f)

$f(x)=(x^2+1)(3-2x^2)$

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Answer.

$f'(x)=-8x^3+2x$

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(g)

$\frac{d}{dx}((x+1)(3x+2))$

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Answer.

$6x + 5$

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(h)

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

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Answer.

$f'(x) = x^2(5x^2+8x+9)$

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(i)

$f(x) = (1-x^2)(2-x^3)$

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Answer.

$f'(x) = 5x^4-3x^2-4x$

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(j)

$f(x) = (3x^3+4)(1-2x^3)$

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Answer.

$f'(x) = -36x^5-15x^2$

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(k)

$f(x) = 3x^4(2x^2-1)$

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Answer.

$f'(x) = 12x^3(3x^2-1)$

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(l)

$f(x) = (x-1)(x^3+x^2+x+1)$

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Answer.

$f'(x) = 4x^3$

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(m)

$f(x) = (5x^7+5x)(6x^3+3x^2+3)$

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Answer.

$f'(x) = 300x^9 + 135x^8 + 105x^6 + 120x^3 + 45x^2 + 15$

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Exercise Group 3.1.6. Differentiate and Simplify (Radicals).

Find each derivative and simplify.

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(a)

$\frac{d}{dx}(x^2(2\sqrt{x} + 1))$

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Answer.

$5x^{3/2} + 2x$

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(b)

$f(x) = \sqrt[3]{x}(1-x)$

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Answer.

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

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(c)

$f(x) = \sqrt{x}(x-2\sqrt{x}+2)$

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Answer.

$f'(x) = \frac{3}{2}\sqrt{x} - 2 + \frac{1}{\sqrt{x}}$

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(d)

$y = (x-x^2)\brac{2x-x^{4/3}}$

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Answer.

$y' = 4x - 6x^2 - \frac{7}{3}x^{4/3} + \frac{10}{3}x^{7/3}$

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(e)

$f(x) = \sqrt{x}(2-x^2+5x^4)$

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Answer.

$f'(x) = \frac{1}{\sqrt{x}} - \frac{5}{2}x^{3/2} + \frac{45}{2}x^{7/2}$

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(f)

$y = (x-\sqrt{x})(x^2+\sqrt{x})$

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Answer.

$y' = 3x^2 - \frac{5}{2}x^{3/2} + \frac{3}{2}\sqrt{x} - 1$

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Exercise Group 3.1.7. Differentiate and Simplify (More Radicals).

Find each derivative and simplify.

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(a)

$y = (2-\sqrt{x})(1+\sqrt{x}+3x)$

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Answer.

$y' = \frac{1+10\sqrt{x}-9x}{2\sqrt{x}}$

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(b)

$y = (2\sqrt{x}-1)(4x+1)^{-1}$

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Answer.

$y' = \frac{-4x+4\sqrt{x}+1}{\sqrt{x}(4x+1)^2}$

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(c)

$y = (2-3\sqrt{x})(4-\sqrt{x})$

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Answer.

$y' = 3 - \frac{7}{\sqrt{x}}$

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Exercise Group 3.1.8. Differentiate with Constants.

Find each derivative and simplify.

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(a)

$f(x) = (ax+b)(cx^2-d)\text{,}$ where $a, b, c, d$ are constants

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Answer.

$f'(x) = 3acx^2+2bcx-ad$

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(b)

$f(x)=\sqrt{x}(a+bx)\text{,}$ where $a$ and $b$ are constants

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Answer.

$f'(x)=\frac{a+3bx}{2\sqrt{x}}$

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Exercise Group 3.1.9. Negative Exponents.

Find each derivative.

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(a)

$f(x) = \brac{\frac{1}{x^2}-\frac{3}{x^4}}(x+5x^3)$

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Answer.

$f'(x)=5+\frac{14}{x^2}+\frac{9}{x^4}$

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(b)

$y = (x^3-2x)(x^{-4}+x^{-2})$

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Answer.

$y'=1+\frac{1}{x^2}+\frac{6}{x^4}$

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(c)

$y = x^{-2}(x^3-3x^2+6)$

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Answer.

$y' = 1 - 12x^{-3}$

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(d)

$f(x) = (6+x^{-2})(8x^{10}-5x^3)$

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Answer.

$f'(x) = -2x^{-3}(8x^{10}-5x^3)+(6+x^{-2})(80x^9-15x^2)$

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(e)

$f(x) = \brac{1+\frac{1}{x^2}}(x^2+1)$

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Answer.

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

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(f)

$y = (x^2+1)\brac{x+5+\frac{1}{x}}$

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Answer.

$y' = 3x^2+10x+2-\frac{1}{x^2}$

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(g)

$y = (1+x^2)(x^{3/4}-x^{-3})$

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Answer.

$y' = \frac{11}{4}x^{7/4}+\frac{3}{4}x^{-1/4}+x^{-2}+3x^{-4}$

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(h)

$f(x) = \frac{1}{x^2} - \frac{3}{x^4}\text{.}$ Hint: Rewrite as $x^{-2} - 3x^{-4}$ and use power rule.

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Answer.

$f'(x)=-2x^{-3}+12x^{-5}$

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(i)

$f(x) = x^2(2+x^{-3})$

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Answer.

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

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Checkpoint 3.1.7. Slope with Negative Exponents.

Find the slope of $y = x^{-5}(1+x^{-1})$ at $x=1\text{.}$

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Answer.

$-11$

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Checkpoint 3.1.8. Derivative Two Ways.

If $f(x) = (6x^4-3x^2+1)(2-x^3)\text{,}$ find $f'(1)$ two ways: (a) by using the Product Rule, and (b) by expanding $f(x)$ first.

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Answer.

$f'(1) = 6$

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Exercise Group 3.1.10. Derivative at a Point.

Find each derivative at the given point.

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(a)

$y = (1+x-2x^2)(3x^3+x-1)\text{,}$ $x = 1$

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Answer.

$-9$

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(b)

$y = (2-3\sqrt{x})(4-\sqrt{x})\text{,}$ $x = 4$

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Answer.

$-\frac{1}{2}$

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Checkpoint 3.1.9. Horizontal Tangent (Linear Factors).

Find the points where the tangent to $y=2(x-29)(x+1)$ is horizontal.

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Answer.

$(14,-450)$

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Checkpoint 3.1.10. Horizontal Tangent (Repeated Factor).

Find the points where the tangent to $y=(x^2+2x+1)(x^2+2x+1)$ is horizontal.

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Answer.

$(-1,0)$

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Subsection 3.1.2 Extended Product Rule (3 or More Factors)

The product rule can be extended to 3 functions multiplied together. It turns out that,

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\begin{equation*} \boxed{(fgh)' = f' gh + fg' h + fgh'} \end{equation*}

Observe the pattern: there are 3 terms, and in each term, exactly one of the functions is differentiated, and the other two are left unchanged.

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Remark 3.1.11.

This rule comes from using the product rule twice. First, we can think of $fgh$ as $(fg)h$ (the function $fg$ together, multiplied by $h$). Then, applying the product rule to $(fg)h\text{,}$ we get,

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\begin{equation*} (fgh)' = (fg)' \cdot h + (fg) \cdot h' \end{equation*}

Then, for $(fg)'\text{,}$ we can apply the product rule again, to get,

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\begin{align*} \amp= (f'g + fg') \cdot h + (fg) \cdot h'\\ \amp= f'gh + fg'h + fgh' \amp\amp \text{distributing} \end{align*}

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Exercise Group 3.1.11. Three-Factor Product Rule.

Find each derivative, using the product rule for 3 functions, and simplify.

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(a)

$f(x) = (x+1)(x+2)(x+3)$

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Answer.

$f'(x) = 3x^2+12x+11$

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(b)

$y = \sqrt{x}(3x+5)(6x^2-5x+1)$

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Answer.

$y' = 63x^{5/2} + \frac{75}{2}x^{3/2} - 33\sqrt{x} + \frac{5}{2\sqrt{x}}$

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Checkpoint 3.1.12. Three-Factor Derivative at a Point.

Find the derivative of $y=x(5x-2)(5x+2)$ at $x=3\text{.}$

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Answer.

$671$

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Subsection 3.1.3 Product Rule with Given Function Values

Exercise Group 3.1.12. Given Function Values.

Find the indicated derivative using the given function values.

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(a)

$f(5) = 1\text{,}$ $f'(5) = 6\text{,}$ $g(5) = -3\text{,}$ $g'(5) = 2\text{.}$ Find $(fg)'(5)\text{.}$

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Hint.

$(fg)'(5) = f'(5)g(5) + f(5)g'(5) = 6(-3) + 1(2)$

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Answer.

$-16$

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(b)

$f(4) = 2\text{,}$ $g(4) = 5\text{,}$ $f'(4) = 6\text{,}$ $g'(4) = -3\text{.}$ Find $(fg)'(4)\text{.}$

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Hint.

$f'(4)g(4) + f(4)g'(4) = 6(5) + 2(-3)$

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Answer.

$24$

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(c)

$f(1) = 5\text{,}$ $f'(1) = 4\text{,}$ $g(1) = 2\text{,}$ $g'(1) = 3\text{.}$ Find $\frac{d}{dx}(f(x)g(x))\big|_{x=1}\text{.}$

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Hint.

$f'(1)g(1) + f(1)g'(1) = 4(2)+5(3)$

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Answer.

$23$

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(d)

$f(3) = 2\text{,}$ $f'(3) = 5\text{,}$ $g(3) = 2\text{,}$ $g'(3) = -10\text{.}$ Find $p'(3)$ where $p(x) = f(x)g(x)\text{.}$

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Hint.

$f'(3)g(3)+f(3)g'(3) = 5(2)+2(-10)$

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Answer.

$-10$

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(e)

$f(2) = 3\text{,}$ $f'(2) = 5\text{,}$ $g(2) = -1\text{,}$ $g'(2) = -4\text{.}$ Find $(fg)'(2)\text{.}$

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Hint.

$(fg)'(2) = f'(2)g(2)+f(2)g'(2) = 5(-1)+3(-4)$

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Answer.

$-17$

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Checkpoint 3.1.13. Tangent Line from Given Values.

Find the equation of the tangent line to the graph of $F(x)=f(x) g(x)$ at $x=1\text{,}$ given that $f(1) = 2\text{,}$ $f'(1) = -3\text{,}$ $g(1) = 4\text{,}$ $g'(1) = -2\text{.}$

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Hint.

$F(1) = 8\text{,}$ $F'(1) = (-3)(4)+(2)(-2) = -16\text{.}$

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Answer.

$y = -16x + 24$

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Subsection 3.1.4 Advanced Examples

Checkpoint 3.1.14. Derivative with Unknown Function.

If $f$ is a differentiable function, find an expression for the derivative of each of the following functions.

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(a)

$y = xf(x)$

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Answer.

$y' = f(x) + xf'(x)$

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(b)

$y = x^2 f(x)$

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Answer.

$y' = 2xf(x) + x^2 f'(x)$

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(c)

$y = \sqrt{x}f(x)$

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Answer.

$y' = \frac{f(x)}{2\sqrt{x}}+\sqrt{x}f'(x)$

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(d)

$y = x^c f(x)$

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Answer.

$y' = cx^{c-1}f(x)+x^c f'(x)$

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Checkpoint 3.1.15. Slope of xf(x).

Find the slope of the tangent line to $y = xf(x)$ at $(1,2)\text{,}$ given $f(1) = 2\text{,}$ $f'(1) = 2\text{.}$

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Hint.

$y' = f(x)+xf'(x)\text{.}$

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Answer.

$4$

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Checkpoint 3.1.16. Tangent to xf(x).

Given that $f(3) = 4\text{,}$ $f'(3) = -2\text{,}$ and $g(x) = xf(x)\text{.}$ Find the tangent to $g$ at $x = 3\text{.}$

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Hint.

$g(3) = 12\text{,}$ $g'(3) = 4 + 3(-2) = -2\text{.}$

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Answer.

$y = -2x + 18$

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Checkpoint 3.1.17. Tangent to x²f(x).

Given that $f(2) = 2\text{,}$ $f'(2) = 3\text{,}$ $g(x) = x^2 f(x)\text{.}$ Find the tangent to $g$ at $x = 2\text{.}$

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Hint.

$g(2) = 8\text{,}$ $g'(2) = 4(2) + 4(3) = 20\text{.}$

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Answer.

$y = 20x - 32$

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Checkpoint 3.1.18. Tangent Using Given Tangent.

The tangent to $h(x)$ at $x = 4$ is $y = -3x + 14\text{.}$ Find the tangent to $y = (x^2-3x)h(x)$ at $x = 4\text{.}$

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Hint.

$h(4) = 2\text{,}$ $h'(4) = -3\text{;}$ $y'(4) = 5(2)+4(-3) = -2\text{,}$ point $(4,8)\text{.}$

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Answer.

$y = -2x+16$

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Checkpoint 3.1.19. Second Derivative.

Given that $f(2) = 10$ and $f'(x) = x^2 f(x)$ for all $x\text{,}$ find $f''(2)\text{.}$

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Hint.

$f''(x) = 2xf(x) + x^2 f'(x)\text{,}$ so $f''(2) = 4(10) + 4(40)\text{.}$

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Answer.

$200$

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Checkpoint 3.1.20. Tangent from Two Tangent Lines.

The tangent to $f$ at $x = 2$ is $y = 4x+1$ and the tangent to $g$ at $x = 2$ is $y = 3x-2\text{.}$ Find the tangent to $y = f(x)g(x)$ at $x = 2\text{.}$

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Hint.

$f(2) = 9\text{,}$ $f'(2) = 4\text{,}$ $g(2) = 4\text{,}$ $g'(2) = 3\text{;}$ $y'(2) = 43\text{,}$ point $(2, 36)\text{.}$

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Answer.

$y = 43x - 50$

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Checkpoint 3.1.21. Derivative of [f(x)]².

(a)

Show that $\frac{d}{dx}[f(x)]^2 = 2f(x)f'(x)\text{,}$ using the product rule with $g=f\text{.}$

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(b)

Use part (a) to differentiate $y = (2+5x-x^3)^2\text{.}$

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Answer.

$y' = 2(2+5x-x^3)(5-3x^2)$

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Checkpoint 3.1.22. Derivative of [f(x)]³.

Show that $\frac{d}{dx}[f(x)]^3 = 3[f(x)]^2 f'(x)\text{,}$ using the product rule for 3 functions.

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Answer.

$(fff)' = f'ff+ff'f+fff' = 3f^2 f'$

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Remark 3.1.23.

In fact, the pattern for the derivative of a product of 3 functions continues for any number of functions. For example, for 4 functions, say, $f, g, h, k\text{,}$ we have,

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\begin{equation*} (fghk)' = f'ghk + fg'hk + fgh'k + fghk' \end{equation*}

In general, for $n$ functions, say, $f_1(x), f_2(x), \dots, f_n(x)\text{,}$

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\begin{align*} \frac{d}{dx} \brac{f_1(x) f_2(x) \cdots f_n(x)} \amp= f_1'(x)f_2(x)\cdots f_n(x) + f_1(x)f_2'(x)\cdots f_n(x) + \cdots + f_1(x)f_2(x)\cdots f_n'(x) \end{align*}

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Checkpoint 3.1.24. f’(0) for n-Factor Product.

If $f(x)=(1+x)(1+2x)(1+3x)\cdots(1+nx)\text{,}$ find $f'(0)\text{.}$

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Hint.

Use the general product rule, and plug in $x=0\text{.}$

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Answer.

$f'(0)=1+2+\cdots+n\text{,}$ which is the sum of the first $n$ natural numbers, which is equal to $\frac{n(n+1)}{2}\text{.}$

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