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Notes

Quotient Rule (Formal)

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

Quotient Rule (Easy to Remember)

\(\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

Problems

Use the quotient rule to find the derivative of the following.

\(\textbf{1)}\) \(f(x)=\displaystyle\frac{x}{x^3+1}\)

Show Answer

The answer is \(f'(x)=\displaystyle\frac{1-2x^3}{(x^3+1)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x}{x^3+1}\right]=\frac{\left(x^3+1\right) \cdot \frac{d}{dx} (x)\,-\,(x) \cdot \frac{d}{dx} \left(x^3+1\right)}{\left(x^3+1\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x}{x^3+1}\right]=\frac{\left(x^3+1\right) \cdot (1)\,-\,(x) \cdot \left(3x^2\right)}{\left(x^3+1\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x}{x^3+1}\right]=\frac{x^3+1-3x^3}{\left(x^3+1\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x}{x^3+1}\right]=\frac{-2x^3+1}{\left(x^3+1\right)^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{1-2x^3}{(x^3+1)^2}\)

\(\textbf{2)}\) \(f(x)=\displaystyle\frac{\sqrt{x}}{x^2-2}\)

Show Answer

The answer is \(f'(x)=-\displaystyle\frac{3x^2+2}{2\sqrt{x}(x^2-2)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\left(x^2-2\right) \cdot \frac{d}{dx} (\sqrt{x})\,-\,(\sqrt{x}) \cdot \frac{d}{dx} \left(x^2-2\right)}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\left(x^2-2\right) \cdot \frac{1}{2\sqrt{x}} \,-\,(\sqrt{x}) \cdot \left(2x\right)}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\frac{x^2-2}{2\sqrt{x}} \,-\,(2x\sqrt{x})}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\frac{x^2-2}{2\sqrt{x}} \,-\,(2x\sqrt{x}) \cdot \frac{2\sqrt{x}}{2\sqrt{x}}}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\frac{x^2-2}{2\sqrt{x}} \,-\, \frac{4x^2}{2\sqrt{x}}}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\frac{x^2-2-4x^2}{2\sqrt{x}}}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=\frac{\frac{-3x^2-2}{2\sqrt{x}}}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\sqrt{x}}{x^2-2}\right]=-\frac{3x^2+2}{2\sqrt{x}} \cdot \frac{1}{\left(x^2-2\right)^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=-\displaystyle\frac{3x^2+2}{2\sqrt{x}(x^2-2)^2}\)

\(\textbf{3)}\) \(f(x)=\displaystyle\frac{\cos ⁡ x}{x^2}\)

Show Answer

The answer is \(f'(x)=-\displaystyle\frac{x \sin x + 2 \cos x}{x^3}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\cos ⁡ x}{x^2}\right]=\frac{\left(x^2\right) \cdot \frac{d}{dx} (\cos ⁡ x)\,-\,(\cos ⁡ x) \cdot \frac{d}{dx} \left(x^2\right)}{\left(x^2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\cos ⁡ x}{x^2}\right]=\frac{\left(x^2\right) \cdot (-\sin ⁡ x)\,-\,(\cos ⁡ x) \cdot \left(2x\right)}{\left(x^2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\cos ⁡ x}{x^2}\right]=\frac{-x^2 \sin (⁡x)\,-\,2x \cos ⁡ (x) }{x^4}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{\cos ⁡ x}{x^2}\right]=\frac{-x \left(x \sin (⁡x)\,+\,2 \cos ⁡ (x)\right) }{x \left(x^3\right)}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=-\displaystyle\frac{x \sin x + 2 \cos x}{x^3}\)

\(\textbf{4)}\) \(f(x)=\displaystyle\frac{x+4}{x-2}\)

Show Answer

The answer is \(f'(x)=-\displaystyle\frac{6}{(x-2)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x+4}{x-2}\right]=\frac{\left(x-2\right) \cdot \frac{d}{dx} (x+4)\,-\,(x+4) \cdot \frac{d}{dx} \left(x-2\right)}{\left(x-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x+4}{x-2}\right]=\frac{\left(x-2\right) \cdot (1)\,-\,(x+4) \cdot \left(1\right)}{\left(x-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x+4}{x-2}\right]=\frac{\left(x-2\right) \,-\,(x+4)}{\left(x-2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x+4}{x-2}\right]=\frac{x-2-x-4}{\left(x-2\right)^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=-\displaystyle\frac{6}{(x-2)^2}\)

\(\textbf{5)}\) \(f(x)=\displaystyle\frac{3x-4}{x^2+3}\)

Show Answer

The answer is \(f'(x)=-\displaystyle\frac{3x^2-8x-9}{(x^2+3)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x-4}{x^2+3}\right]=\frac{\left(x^2+3\right) \cdot \frac{d}{dx} (3x-4)\,-\,(3x-4) \cdot \frac{d}{dx} \left(x^2+3\right)}{\left(x^2+3\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x-4}{x^2+3}\right]=\frac{\left(x^2+3\right) \cdot (3)\,-\,(3x-4) \cdot \left(2x\right)}{\left(x^2+3\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x-4}{x^2+3}\right]=\frac{\left(3x^2+9\right) \,-\,(6x^2-8x)}{\left(x^2+3\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x-4}{x^2+3}\right]=\frac{3x^2+9-6x^2+8x}{\left(x^2+3\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x-4}{x^2+3}\right]=\frac{-3x^2+8x+9}{\left(x^2+3\right)^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=-\displaystyle\frac{3x^2-8x-9}{(x^2+3)^2}\)

\(\textbf{6)}\) \(f(x)=\displaystyle\frac{e^x}{x}\)

Show Answer

The answer is \(f'(x)=\displaystyle\frac{e^x(x-1)}{x^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x}\right]=\frac{\left(x\right) \cdot \frac{d}{dx} (e^x)\,-\,(e^x) \cdot \frac{d}{dx} \left(x\right)}{\left(x\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x}\right]=\frac{\left(x\right) \cdot (e^x)\,-\,(e^x) \cdot \left(1\right)}{\left(x\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x}\right]=\frac{x e^x-e^x }{x^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{e^x(x-1)}{x^2}\)

\(\textbf{7)}\) \(f(x)=\displaystyle\frac{x^2}{x+1}\)

Show Answer

The answer is \(f'(x)=\displaystyle\frac{x(x+2)}{(x+1)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x^2}{x+1}\right]=\frac{\left(x+1\right) \cdot \frac{d}{dx} (x^2)\,-\,(x^2) \cdot \frac{d}{dx} \left(x+1\right)}{\left(x+1\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x^2}{x+1}\right]=\frac{\left(x+1\right) \cdot (2x)\,-\,(x^2) \cdot \left(1\right)}{\left(x+1\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x^2}{x+1}\right]=\frac{2x^2+2x-x^2}{\left(x+1\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{x^2}{x+1}\right]=\frac{x^2+2x}{\left(x+1\right)^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{x(x+2)}{(x+1)^2}\)

\(\textbf{8)}\) \(f(x)=\displaystyle\frac{e^x}{x+2}\)

Show Answer

The answer is \(f'(x)=\displaystyle\frac{e^x(x+1)}{(x+2)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x+2}\right]=\frac{\left(x+2\right) \cdot \frac{d}{dx} (e^x)\,-\,(e^x) \cdot \frac{d}{dx} \left(x+2\right)}{\left(x+2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x+2}\right]=\frac{\left(x+2\right) \cdot (e^x)\,-\,(e^x) \cdot \left(1\right)}{\left(x+2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x+2}\right]=\frac{xe^x + 2e^x \,-\,e^x}{\left(x+2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{e^x}{x+2}\right]=\frac{xe^x + e^x}{\left(x+2\right)^2}\)

\(\,\,\,\,\,\,\)The answer is \(f'(x)=\displaystyle\frac{e^x(x+1)}{(x+2)^2}\)

\(\textbf{9)}\) \(f(x)=\displaystyle\frac{3x^3-2x}{1+3x}\)

Show Answer

The answer is \(f'(x)=\displaystyle\frac{18x^3+9x^2-2}{\left(1+3x\right)^2}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{hi}{lo}\right]=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x^3-2x}{1+3x}\right]=\frac{\left(1+3x\right) \cdot \frac{d}{dx} (3x^3-2x)\,-\,(3x^3-2x) \cdot \frac{d}{dx} \left(1+3x\right)}{\left(1+3x\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x^3-2x}{1+3x}\right]=\frac{\left(1+3x\right) \cdot \left(9x^2-2\right)\,-\,\left(3x^3-2x\right) \cdot \left(3\right)}{\left(1+3x\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x^3-2x}{1+3x}\right]=\frac{\left(1+3x\right)\left(9x^2-2\right)\,-\,\left(3x^3-2x\right)\left(3\right)}{\left(1+3x\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x^3-2x}{1+3x}\right]=\frac{9x^2-2+27x^3-6x-9x^3+6x}{\left(1+3x\right)^2}\)

\(\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\frac{3x^3-2x}{1+3x}\right]=\frac{18x^3+9x^2-2}{\left(1+3x\right)^2}\)

\(\textbf{10)}\) Find \((f/g)'(4)\) where \(f(4)=3\), \(f'(4)=6\), \(g(4)=2\), \(g'(4)=0\)

Show Answer

The answer is \((f/g)'(4)=3\)

Show Work

\(\,\,\,\,\,\,\displaystyle (f/g)'(4)=\frac{lo \cdot d hi\,-\,hi \cdot d lo}{lo^2}\)

\(\,\,\,\,\,\,\displaystyle (f/g)'(4)=\frac{g(4) \cdot f'(4)\,-\,f(4) \cdot g'(4)}{\left(g(4)\right)^2}\)

\(\,\,\,\,\,\,\displaystyle (f/g)'(4)=\frac{2 \cdot 6\,-\,3 \cdot 0}{\left(2\right)^2}\)

\(\,\,\,\,\,\,\displaystyle (f/g)'(4)=\frac{12\,-\,0}{4}\)

\(\,\,\,\,\,\,\)The answer is \((f/g)'(4)=3\)

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Definition of Derivative}\)
\(\,\,\,\,\,\,\,\, \displaystyle \lim_{\Delta x\to 0} \frac{f(x+ \Delta x)-f(x)}{\Delta x} \)

\(\bullet\text{ Equation of the Tangent Line}\)
\(\,\,\,\,\,\,\,\,f(x)=x^3+3x^2−x \text{ at the point } (2,18)\)

\(\bullet\text{ Derivatives- Constant Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(c)=0\)

\(\bullet\text{ Derivatives- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(x^n)=nx^{n-1}\)

\(\bullet\text{ Derivatives- Constant Multiple Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}(cf(x))=cf'(x)\)

\(\bullet\text{ Derivatives- Sum and Difference Rules}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \pm g(x)]=f'(x) \pm g'(x)\)

\(\bullet\text{ Derivatives- Sin and Cos}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}sin(x)=cos(x)\)

\(\bullet\text{ Derivatives- Product Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(x) \cdot g(x)]=f(x) \cdot g'(x)+f'(x) \cdot g(x)\)

\(\bullet\text{ Derivatives- Quotient Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}\left[\displaystyle\frac{f(x)}{g(x)}\right]=\displaystyle\frac{g(x) \cdot f'(x)-f(x) \cdot g'(x)}{[g(x)]^2}\)

\(\bullet\text{ Derivatives- Chain Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[f(g(x))]= f'(g(x)) \cdot g'(x)\)

\(\bullet\text{ Derivatives- ln(x)}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{d}{dx}[ln(x)]= \displaystyle \frac{1}{x}\)

\(\bullet\text{ Implicit Differentiation}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Horizontal Tangent Line}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Mean Value Theorem}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Related Rates}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Increasing and Decreasing Intervals}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Intervals of concave up and down}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Inflection Points}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Graph of f(x), f'(x) and f”(x)}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Newton’s Method}\)
\(\,\,\,\,\,\,\,\,x_{n+1}=x_n – \displaystyle \frac{f(x_n)}{f'(x_n)}\)

In Summary

The Quotient Rule is a mathematical formula that allows us to differentiate a function that is the quotient of two other functions. In other words, it allows us to find the derivative of a function that is expressed as the division of one function by another.

The formal definition of the Quotient Rule states that if we have two functions, \(f(x)\) and \(g(x)\), the derivative of their quotient, \(h(x) = f(x)/g(x)\), is given by the formula: \(h'(x) = \displaystyle\frac{g(x) \cdot f'(x) – f(x) \cdot g'(x)}{g(x)^2}\).

The Quotient Rule is typically taught early on in a Calculus class. It is one of the most popular rules related to taking derivatives.

One common mistake when using the Quotient Rule is to use it for something like \(f(x)=\frac{3}{(x+5)^2}\). Since the 3 in the denominator is a constant, the quotient rule is not the easiest way to find the derivative. You can rewrite the function as \(f(x)=3(x+5)^{-2}\) and use the power rule for derivatives. While both methods will work, this second method should be much quicker.

A fun fact about the Quotient Rule is that it was developed by mathematician and philosopher Gottfried Wilhelm Leibniz, who is known for his contributions to calculus and the development of the notation we use today. Leibniz is often credited as the co-inventor of calculus, along with Sir Isaac Newton.

The Quotient Rule, the Power Rule, the Chain Rule, and the Product Rule are the first methods you learn to find the derivatives of functions.

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