# Quotient Rule

This discussion will focus on the Quotient Rule of Differentiation. This rule states that:

The derivative of the quotient of two functions is equal to the denominator multiplied by the derivative of the numerator minus the numerator multiplied by the derivative of the denominator, all divided by the denominator squared.

QUOTIENT RULE OF DIFFERENTIATION:

$\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^{2}}$

Let's work some examples

To work these examples requires the use of various differentiation rules. If you are not familiar with a rule go to the associated topic for a review.

$\frac{{x}^{3}-5x+8}{{x}^{2}-15}$
 Step 1: Simplify the numerator and denominator. Both the numerator and denominator are already simplified. $\frac{{x}^{3}-5x+8}{{x}^{2}-15}$ Step 2: Apply the quotient rule. $\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^{2}}$ $\frac{d}{dx}\left[\frac{{x}^{3}-5x+8}{{x}^{2}-15}\right]$ $\frac{\left[\left({x}^{2}-15\right)\frac{d}{dx}\left({x}^{3}-5x+8\right)\right]-\left[\left({x}^{3}-5x+8\right)\frac{d}{dx}\left({x}^{2}-15\right)\right]}{{\left({x}^{2}-15\right)}^{2}}$ Step 3: Calculate the two derivatives. $\frac{d}{dx}\left({x}^{3}-5x+8\right)$ Original $\frac{d}{dx}{x}^{3}-\frac{d}{dx}5x+\frac{d}{dx}8$    Sum/Diff Rule $\frac{d}{dx}{x}^{3}-5\frac{d}{dx}x+\frac{d}{dx}8$    Constant Multiple 3x2 - 5(1) + 0 Power & Constant $3{x}^{2}-5$ ______________________________ $\frac{d}{dx}\left({x}^{2}-15\right)$ Original $\frac{d}{dx}{x}^{2}-\frac{d}{dx}15$   Difference Rule 2x1 - 0 Power & Constant $2x$ Step 4: Substitute the derivatives & simplify. $\frac{\left[\left({x}^{2}-15\right)\left(3{x}^{2}-5\right)\right]-\left[\left({x}^{3}-5x+8\right)\left(2x\right)\right]}{{\left({x}^{2}-15\right)}^{2}}$ $\frac{\left(3{x}^{4}-5{x}^{2}-45{x}^{2}+75\right)-\left(2{x}^{4}-10{x}^{2}+16x\right)}{{\left({x}^{2}-15\right)}^{2}}$ $\frac{\left(3{x}^{4}-50{x}^{2}+75\right)+\left(-2{x}^{4}+10{x}^{2}-16x\right)}{{\left({x}^{2}-15\right)}^{2}}$ $\frac{{x}^{4}-40{x}^{2}-16x+75}{{\left({x}^{2}-15\right)}^{2}}$
Example 1:      $\frac{3{e}^{x}}{\sqrt{x}}$
 Step 1: Simplify the numerator and denominator. Both the numerator and denominator are already simplified. $\frac{3{e}^{x}}{\sqrt{x}}$ Step 2: Apply the quotient rule. $\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^{2}}$ $\frac{d}{dx}\left[\frac{3{e}^{x}}{\sqrt{x}}\right]$ $\frac{\left[\left(\sqrt{x}\right)\frac{d}{dx}\left(3{e}^{x}\right)\right]-\left[\left(3{e}^{x}\right)\frac{d}{dx}\left(\sqrt{x}\right)\right]}{{\left(\sqrt{x}\right)}^{2}}$ Step 3: Calculate the two derivatives. $\frac{d}{dx}\left(3{e}^{x}\right)$      Original $3\frac{d}{dx}{e}^{x}$    Constant Mult. 3ex        Natural Exp. $3{e}^{x}$ ______________________________ $\frac{d}{dx}\left(\sqrt{x}\right)$    Original $\frac{1}{2}{x}^{-\frac{1}{2}}$       Power $\frac{1}{2\sqrt{x}}$ Step 4: Substitute the derivatives & simplify. $\frac{3{e}^{x}\left(\sqrt{x}-\frac{1}{2\sqrt{x}}\right)}{x}$
Example 2:      $\frac{2{x}^{3}}{1-3{x}^{2}}$
 Step 1: Simplify the numerator and denominator. Both the numerator and denominator are already simplified. $\frac{2{x}^{3}}{1-3{x}^{2}}$ Step 2: Apply the quotient rule. $\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]-f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]}{{\left[g\left(x\right)\right]}^{2}}$ $\frac{d}{dx}\left[\frac{2{x}^{3}}{1-3{x}^{2}}\right]$ $\frac{\left[\left(1-3{x}^{2}\right)\frac{d}{dx}\left(2{x}^{3}\right)\right]-\left[\left(2{x}^{3}\right)\frac{d}{dx}\left(1-3{x}^{2}\right)\right]}{{\left(1-3{x}^{2}\right)}^{2}}$ Step 3: Calculate the two derivatives. $\frac{d}{dx}\left(2{x}^{3}\right)$    Original $2\frac{d}{dx}{x}^{3}$    Constant Multiple 2(3x2)      Power $6{x}^{2}$ ________________________ $\frac{d}{dx}\left(1-3{x}^{2}\right)$ Original $\frac{d}{dx}1-\frac{d}{dx}3{x}^{2}$ Difference Rule $\frac{d}{dx}1-3\frac{d}{dx}{x}^{2}$ Constant Multiple 0 - 3(2x) Constant/Power $-6x$ Step 4: Substitute the derivatives & simplify. $\frac{\left[\left(1-3{x}^{2}\right){6}{x}^{2}\right]-\left[\left(2{x}^{3}\right)\left(-6x\right)\right]}{{\left(1-3{x}^{2}\right)}^{2}}$ $\frac{6{x}^{2}-18{x}^{4}+12{x}^{4}}{{\left(1-3{x}^{2}\right)}^{2}}$ $\frac{6{x}^{2}-6{x}^{4}}{{\left(1-3{x}^{2}\right)}^{2}}$

 Related Links: Math algebra Chain Rule Trigonometry Differentiation Rules Calculus Topics

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