Product Rule Differentiation


The product rule differentiation is used in differential calculus to help calculating the derivative of products of functions. The formula for the product rule differentiation is written for the product of two or more functions.
If $u(x)$ and $v(x)$ are two functions of $x$ and $f(x)=u(x)v(x)$, then
$$f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)$$
Alternatively if $y=u \times v$ then
$$\dfrac{dy}{dx} = \dfrac{du}{dx} \times v + u \times \dfrac{dv}{dx}$$
Expressions like $(x^2-1)(3x^3+2x-1)$, $x^5(x^2+2x-1)^3$ and $(x+1)^4(2x^3-2)^5$ are called products because they represent the product of one function by another.

Note that applying the chain rule needs to be considered at any time. This is where most students forget when they applying chain rule and resulted in silly mistakes a lot.
No matter what rule you apply, the chain rule should be considered to apply in any case for differentiating. Of course, more than two tiers of chain rule are quite common in some complicated questions.
Therefore, if you are serious about applying the chain rule while differentiating using either quotient rule or product rule differentiation, then you will definitely be likely to get better results than others.

Watch out! The chain rule goes everywhere!

Example 1

Use the product rule to differentiate $f(x)=x(2x+3)^4$, simplify where possible.

\( \begin{align} \displaystyle
f^{\prime}(x) &= x^{\prime} \times (2x+3)^4 + x \times \big[(2x+3)^4\big]^{\prime} \\
&= 1 \times (2x+3)^4 + x \times 4(2x+3)^{4-1} \times (2x+3)’ \\
&= (2x+3)^4 + 4x(2x+3)^3 \times 2 \\
&= (2x+3)^4 + 8x(2x+3)^3 \\
&= (2x+3)^3\big[(2x+3) + 8x\big] \\
&= (2x+3)^3(10x+3) \\
\end{align} \)

Example 2

Use the product rule to differentiate $f(x)=\sqrt{x}(x^2-2)^{10}$.

\( \begin{align} \displaystyle
\sqrt{x} &= x^{\frac{1}{2}} \\
\sqrt{x}^{\prime} &= \dfrac{1}{2}x^{\frac{1}{2}-1} \\
&= \dfrac{1}{2}x^{-\frac{1}{2}} \\
&= \dfrac{1}{2\sqrt{x}} \\
f^{\prime}(x) &= \sqrt{x}^{\prime} \times (x^2-2)^{10} + \sqrt{x} \times \big[(x^2-2)^{10}\big]^{\prime} \\
&= \dfrac{1}{2\sqrt{x}} \times (x^2-2)^{10} + \sqrt{x} \times 10(x^2-2)^{10-1} \times (x^2-2)^{\prime} \\
&= \dfrac{1}{2\sqrt{x}} \times (x^2-2)^{10} + \sqrt{x} \times 10(x^2-2)^{9} \times 2x \\
&= \dfrac{(x^2-2)^{10}}{2\sqrt{x}} + 20x\sqrt{x}(x^2-2)^9 \\
\end{align} \)

Example 3

Use the product rule to differentiate $f(x)=(x^2+1)^5(x^3-1)^4$, and simplify where possible.

\( \begin{align} \displaystyle
f^{\prime}(x) &= \big[(x^2+1)^5\big]^{\prime} \times (x^3-1)^4 + (x^2+1)^5 \times \big[(x^3-1)^4\big]^{\prime} \\
&= 5(x^2+1)^{5-1} \times (x^2+1)^{\prime} \times (x^3-1)^4 + (x^2+1)^5 \times 4(x^3-1)^{4-1} \times (x^3-1)^{\prime} \\
&= 5(x^2+1)^{4} \times 2x \times (x^3-1)^4 + (x^2+1)^5 \times 4(x^3-1)^{3} \times 3x^2 \\
&= 10x(x^2+1)^{4}(x^3-1)^4 + 12x^2(x^2+1)^5(x^3-1)^{3} \\
&= 2x(x^2+1)^{4}(x^3-1)^3\big[5(x^3-1)+6x(x^2+1)\big] \\
&= 2x(x^2+1)^{4}(x^3-1)^3(5x^3-5+6x^3+6x) \\
&= 2x(x^2+1)^{4}(x^3-1)^3(11x^3+6x-5) \\
\end{align} \)

Extension Examples

These Extension Examples require to have some prerequisite skills including;
\( \begin{align} \displaystyle
\dfrac{d}{dx}\sin{x} &= \cos{x} \\
\dfrac{d}{dx}\cos{x} &= -\sin{x} \\
\dfrac{d}{dx}e^x &= e^x \\
\dfrac{d}{dx}\log_e{x} &= \dfrac{1}{x} \\
\end{align} \)

Example 4

Find $\displaystyle \dfrac{dy}{dx}$ of $y=x\sin{x}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= \dfrac{d}{dx}x \times \sin{x} + x \times \dfrac{d}{dx}\sin{x} \\
&= 1 \times \sin{x} + x \times \cos{x} \\
&= \sin{x} + x\cos{x} \\
\end{align} \)

Example 5

Find $\displaystyle \dfrac{dy}{dx}$ of $y=\sin{x}\cos{x}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$ and $\dfrac{d}{dx}\cos{x} = -\sin{x}$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= \dfrac{d}{dx}\sin{x} \times \cos{x} + \sin{x} \times \dfrac{d}{dx}\cos{x} \\
&= \cos{x} \times \cos{x} + \sin{x} \times (-\sin{x}) \\
&= \cos^2{x} – \sin^2{x} \\
\end{align} \)

Example 6

Find $\displaystyle \dfrac{dy}{dx}$ of $y=x^2\log_e{2x}$, known that $\dfrac{d}{dx}\log_e{x} = \dfrac{1}{x}$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= \dfrac{d}{dx}x^2 \times \log_e{2x} + x^2 \times \dfrac{d}{dx}\log_e{2x} \\
&= 2x \times \log_e{2x} + x^2 \times \dfrac{1}{2x} \times \dfrac{d}{dx}(2x) \\
&= 2x\log_e{2x} + x^2 \times \dfrac{1}{2x} \times 2 \\
&= 2x\log_e{2x} + x \\
\end{align} \)

Example 7

Find $\displaystyle \dfrac{dy}{dx}$ of $y=e^{2x}\sin{3x}$, known that $\dfrac{d}{dx}\sin{x} = \cos{x}$ and $\dfrac{d}{dx}e^x = e^x$.

\( \begin{align} \displaystyle
\dfrac{dy}{dx} &= \dfrac{d}{dx}e^{2x} \times \sin{3x} + e^{2x} \times \dfrac{d}{dx}\sin{3x} \\
&= e^{2x} \times \dfrac{d}{dx}(2x) \times \sin{3x} + e^{2x} \times \cos{3x} \times \dfrac{d}{dx}(3x) \\
&= e^{2x} \times 2 \times \sin{3x} + e^{2x} \times \cos{3x} \times 3 \\
&= 3e^{2x}sin{3x} + 3e^{2x}\cos{3x} \\
\end{align} \)


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