2.5 Product Rule

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We compute the derivative of a product.

The Product Rule

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

In words, the derivative of a product is the derivative of the first times the second plus the first times the derivative of the second.

example 1 Find $\frac {d}{dx} [(x^2 + 5x+3)(x^3 - 6x^2 -7)]$

Use the product rule with $f(x) = x^2 + 5x+3$ and $g(x) = x^3 -6x^2 -7.$ With these choices, we have $f'(x) = 2x + 5$ and $g'(x) = 3x^2 - 12x.$ The product rule then gives $\frac {d}{dx} [(x^2 + 5x+3)(x^3 - 6x^2 -7)]\\ =(2x+5)(x^3-6x^2 -7) + (x^2 + 5x+3)(3x^2 -12x).$

(problem 1) Compute $\frac {d}{dx}[(x^2 - 3x +1)(x^3 + 4x^2 -2)]$

Use the Product Rule with $x^2 -3x+1$ and $x^3 +4x^2-2$ .

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

The derivative of $x^2 -3x+1$ is $2x-3$

The derivative of $x^3 + 4x^2 -2$ is $3x^2 +8x$

The derivative of $(x^2 - 3x +1)(x^3 + 4x^2 -2)$ with respect to $x$ is $\answer {(2x-3)(x^3+4x^2-2)+(x^2-3x+1)(3x^2+8x)}$

example 2 Find $\frac {d}{dx} x^3e^x$

Use the product rule with $f(x) = x^3$ and $g(x) = e^x.$ With these choices, we have $f'(x) = 3x^2$ and $g'(x) = e^x.$ The product rule then gives $\frac {d}{dx}x^3e^x = 3x^2e^x + x^3e^x\\ =(x^3 + 3x^2)e^x.$

(problem 2) Compute $\frac {d}{dx} \sqrt x e^x$

Use the Product Rule with $\sqrt x$ and $e^x$ .

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

The derivative of $\sqrt x$ is $\frac {1}{2\sqrt x}$

The derivative of $e^x$ is $e^x$

The derivative of $\sqrt x e^x$ with respect to $x$ is $\answer {(\sqrt x + 1/(2\sqrt x))e^x}$

example 3 Find $h'(x)$ if $h(x) = x^3\sin (x)$ .
We write $h(x)$ as a product: $h(x) = f(x)\cdot g(x)$ with $f(x) = x^3 \quad \text {and} \quad g(x) = \sin (x).$ To use the product rule, we need the derivatives: $f'(x) = 3x^2 \quad \text {and} \quad g'(x) = \cos (x).$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 3x^2\sin(x) + x^3\cos(x). \end{align*}$

Here is a video of the preceding example

(problem 3a) Compute $\frac {d}{dx} x^4\sin (x)$

Use the Product Rule with $x^4$ and $\sin (x)$ .

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

The derivative of $x^4$ is $4x^3$

The derivative of $\sin (x)$ is $\cos (x)$

The derivative of $x^4\sin (x)$ with respect to $x$ is $\answer {4x^3\sin (x) + x^4\cos (x)}$

(problem 3b) Compute $\frac {d}{dx} 2\sqrt x\sin (x)$

Use the Product Rule with $2\sqrt x$ and $\sin (x)$ .

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

The derivative of $2\sqrt x$ is $\frac {1}{\sqrt x}$

The derivative of $\sin (x)$ is $\cos (x)$

The derivative of $2\sqrt x\sin (x)$ with respect to $x$ is $\answer {\frac {\sin (x)}{\sqrt x} + 2\sqrt x\cos (x)}$

example 4 Find $h'(x)$ if $h(x) = 3x^5\cos (x)$ .
We write $h(x)$ as a product: $h(x) = f(x)\cdot g(x)$ with $f(x) = 3x^5 \quad \text {and} \quad g(x) = \cos (x).$ To use the product rule, we need the derivatives $f'(x) = 15x^4 \quad \text {and} \quad g'(x) = -\sin (x).$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x)\\ &= 15x^4 \cos(x) +3x^5(-\sin(x)) \\ &= 15x^4 \cos(x) - 3x^5\sin(x). \end{align*}$

Here is a video of the preceding example

(problem 4a) Compute $\frac {d}{dx} x^5\cos (x)$

Use the Product Rule with $x^5$ and $\cos (x)$ .

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

The derivative of $x^5$ is $5x^4$

The derivative of $\cos (x)$ is $-\sin (x)$

The derivative of $x^5\cos (x)$ with respect to $x$ is $\answer {5x^4\cos (x) - x^5\sin (x)}$

(problem 4b) Compute $\frac {d}{dx} \sqrt [3]x \cos (x)$

Use the Product Rule with $\sqrt [3] x$ and $\cos (x)$ .

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

The derivative of $\sqrt [3]x$ is $\frac {1}{3\sqrt [3]x^2}$

The derivative of $\cos (x)$ is $-\sin (x)$

The derivative of $\sqrt [3]x\cos (x)$ with respect to $x$ is $\answer {-x^{1/3}\sin (x) + (1/3)x^{-2/3}\cos (x)}$

example 5 Find $h'(x)$ if $h(x) = (x^2 - 5x - 2)e^x$ .
We write $h(x)$ as a product: $h(x) = f(x) \cdot g(x)$ with $f(x) = x^2 - 5x -2 \quad \text {and} \quad g(x) = e^x.$ To use the product rule, we need the derivatives: $f'(x) = 2x - 5 \quad \text {and} \quad g'(x)= e^x.$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= (2x - 5)e^x+ (x^2 - 5x - 2)e^x \\ &= (x^2 - 5x - 2 + 2x - 5)e^x \\ &= (x^2 - 3x - 7)e^x. \end{align*}$

Here is a video of the preceding example

(problem 5) Compute $\frac {d}{dx} (x^3 - 5x^2 + 3x -2)e^x$

Use the Product Rule with $x^3 - 5x^2 + 3x -2$ and $e^x$ .

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

The derivative of $x^3 - 5x^2 + 3x -2$ is $3x^2 - 10x + 3$

The derivative of $e^x$ is itself

Factor out $e^x$ and combine like terms

The derivative of $(x^3 - 5x^2 + 3x -2)e^x$ with respect to $x$ is $\answer {(x^3 - 2x^2 - 7x +1)e^x}$

example 6 Find $h'(x)$ if $h(x) = x\ln (x)$ .
We write $h(x)$ as a product: $h(x) = f(x) \cdot g(x)$ with $f(x) = x \quad \text {and} \quad g(x) = \ln (x).$ To use the product rule, we need the derivatives: $f'(x) = 1 \quad \text {and} \quad g'(x)= \frac {1}{x}.$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= 1 \cdot \ln(x) + x\frac{1}{x} \\ &= \ln(x) + 1. \end{align*}$

Here is a video of the preceding example

(problem 6) Compute $\frac {d}{dx} x^3\ln (x)$

Use the Product Rule with $x^3$ and $\ln (x)$ .

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

The derivative of $x^3$ is $3x^2$

The derivative of $\ln (x)$ is $1/x$

The derivative of $x^3\ln (x)$ with respect to $x$ is $\answer {x^2(3 \ln (x) + 1)}$

example 7 Find $h'(x)$ if $h(x) = e^x\sin (x)$ .
We write $h(x)$ as a product: $h(x) = f(x)\cdot g(x)$ with $f(x) = e^x \quad \text {and} \quad g(x) = \sin (x).$ To use the product rule, we need the derivatives: $f'(x) = e^x \quad \text {and} \quad g'(x) = \cos (x).$ We can now write the derivative: $\begin{align*} h'(x) &= f'(x)g(x) + f(x)g'(x) \\ &= e^x \sin(x) + e^x \cos(x) \\ &= e^x[\sin(x) + \cos(x)]. \end{align*}$

Here is a video of the preceding example

(problem 7) Compute $\frac {d}{dx} 3e^x \cos (x)$

Use the Product Rule with $3e^x$ and $\cos (x)$ .

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

The derivative of $3e^x$ is itself

The derivative of $\cos (x)$ is $-\sin (x)$

Factor out $3e^x$

The derivative of $3e^x\cos (x)$ with respect to $x$ is $\answer {3e^x(\cos (x) - \sin (x))}$

Here is a detailed, lecture style video on the Product Rule:

We conclude this section with a derivation of the product rule using the definition of the derivative.

$\begin{align*} \left[f(x)g(x)\right]' &= \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{f(x+h)g(x+h)- f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)}{h}\\[5pt] &= \lim_{h \to 0} \frac{f(x+h)g(x+h)- f(x)g(x+h)}{h} + \frac{f(x)g(x+h) - f(x)g(x)}{h} \\[5pt] &= \lim_{h \to 0} \frac{g(x+h)\left[f(x+h)- f(x)\right]}{h} + \frac{f(x)\left[g(x+h) - g(x)\right]}{h}\\[5pt] &= g(x)f'(x) + f(x)g'(x). \end{align*}$