Product Rule Differentiation

Product Rule Differentiation

The product rule differentiation is used in differential calculus to help calculate 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
$$ \large f^{\prime}(x) = u^{\prime}(x)v(x) + u(x)v^{\prime}(x)$$
Alternatively if $y=u \times v$ then
$$ \large \dfrac{dy}{dx} = \dfrac{du}{dx} \times v + u \times \dfrac{dv}{dx}$$
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 apply the chain rule, which greatly results in silly mistakes.
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 the quotient rule or product rule differentiation, you will likely 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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