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Differentiation: (lesson 2 of 3)

Product and Quotient Rule for Derivatives

Product Rule

Let $f(x)$ and $g(x)$ be differentiable at $x$. Then $h(x) = f(x) g(x)$ is differentiable at $x$ and

$$\color{blue}{(f(x) \cdot g(x))' = f'(x) \cdot g(x) + f(x) \cdot g'(x)}$$

We illustrate product rule with the following examples:

Example 1:

\begin{aligned} (\color{blue}{x^3} \cdot \color{red}{e^x})' &= \color{blue}{(x^3)'} \cdot \color{red}{(e^x)'} + \color{blue}{x^3} \cdot \color{red}{(e^x)'} \\ &= 3 \cdot x^2 \cdot e^x + x^3 \cdot e^x = \\ &= x^2 e^x (3 + x) \end{aligned}

Example 2:

\begin{aligned} (\color{blue}{(x + 2)} \cdot \color{red}{(x^2 - x + 1)})' &= \color{blue}{(x + 2)'} \cdot \color{red}{(x^2 - x + 1)} + \color{blue}{(x + 2)} \cdot \color{red}{(x^2 - x + 1)'} = \\ &= 1 \cdot (x^2 - x + 1) + (x + 2) \cdot (2x - 1) = \\ &= x^2 - x + 1 + x^2 - x + 4x - 2 = \\ &= 2x^2 + 2x - 1 \end{aligned}
 Try yourself $$\color{blue}{[ ( x^2 + 3x - 1 ) \cdot e^x ]' = }$$ $( x^2 - 5x - 2 ) \cdot e^x$ $( x^2 - 5x + 2 ) \cdot e^x$ $( x^2 + 5x - 2 ) \cdot e^x$ $( x^2 + 5x + 2 ) \cdot e^x$

Quotient rule:

Let $f(x)$ and $g(x)$ be differentiable at $x$ with $g(x) \ne 0$. Then $\frac{f(x)}{g(x)}$ is differentiable at $x$ and

$$\color{blue}{\left[\frac{f(x)}{g(x)}\right]' = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}}$$

We illustrate quotient rule with the following examples:

Example 3:Differentiate $y = \frac{x^2}{2 + x}$

Solution 3:

\begin{aligned} \left( \frac{\color{blue}{x^2}}{\color{red}{2 + x}} \right)' &= \frac{\color{blue}{(x^2)'} \cdot \color{red}{(2 + x)} - \color{blue}{x^2} \cdot \color{red}{(2 + x)'}}{\color{red}{(2 + x)^2}} = \\ &= \frac{2x \cdot (2 + x) - x^2 \cdot (0+1)}{(2 + x)^2} = \\ &= \frac{4x + 2x^2 - x^2}{(2 + x)^2} = \\ &= \frac{x^2 + 4x}{(2 + x)^2} \end{aligned}
 Try yourself $$\color{blue}{\left( \frac{3 + x}{x^2} \right)' = }$$ $\frac{x + 6}{x^3}$ $- \frac{x + 6}{x^3}$ $\frac{x + 6}{x^4}$ $-\frac{x + 6}{x^4}$

Example 4:Differentiate $y = \frac{\ln x}{1 + x^2}$

Solution 4:

\begin{aligned} \left( \frac{\color{blue}{\ln x}}{\color{red}{x^2}} \right)' &= \frac{\color{blue}{(\ln x)'} \cdot \color{red}{x^2} - \color{blue}{\ln x} \cdot \color{red}{(x^2)}}{\color{red}{(x^2)^2}} = \\ &= \frac{\frac{1}{x} \cdot x^2 - \ln x \cdot 2x}{x^4} = \\ &= \frac{x - 2 x \ln x}{x^4} = \\ &= \frac{x (1 - 2 \ln x)}{x^4} = \\ &= \frac{1 - 2 \ln x}{x^3} \end{aligned}
 Try yourself $$\color{blue}{\left( \frac{\ln x}{x^2 + x} \right)' = }$$ $\frac{x + 1 - ( 2x - 1 ) \ln x}{{(x^2 + x)}^2}$ $\frac{x - 1 + ( 2x - 1 ) \ln x}{{(x^2 + x)}^2}$ $\frac{x + 1 - ( 2x + 1 ) \ln x}{{(x^2 + x)}^2}$ $\frac{x - 1 + ( 2x + 1 ) \ln x}{{(x^2 + x)}^2}$

Example 5:Differentiate $y = \tan x$

Solution 5:

\begin{aligned} \left( \frac{\color{blue}{\sin x}}{\color{red}{\cos x}} \right)' &= \frac{\color{blue}{( \sin x )'} \cdot \color{red}{\cos x} - \color{blue}{\sin x} \cdot \color{red}{(\cos x)'}}{\color{red}{(\cos x)^2}} = \\ &= \frac{\cos x \cdot \cos x - \sin x \cdot ( - \sin x)}{{\cos}^2 x} = \\ &= \frac{{\cos}^2 x + {\sin}^2 x}{{\cos}^2 x} = \\ &= \frac{1}{{\cos}^2 x} \end{aligned}
 Try yourself $$\color{blue}{( \cot x )' = }$$ $\frac{1}{\cos^2 x}$ $- \frac{1}{\cos^2 x}$ $\frac{1}{\sin^2 x}$ $- \frac{1}{\sin^2 x}$