The Quotient Rule for Differentiation

Core Concepts

Details the Quotient Rule, used for finding the derivative of a rational function or any function expressed as a ratio f(x)/g(x). Includes the formula, a popular mnemonic device, and various examples, including when to avoid the rule.

The Quotient Rule for Differentiation

1 of 14

When must you use the Quotient Rule for differentiation?

It is used to find the derivative of a function that is a ratio of two other functions, $h(x) = \frac{f(x)}{g(x)}$, especially when the denominator $g(x)$ is not a constant.

State the Quotient Rule for differentiation using Lagrange notation.

If $h(x) = \frac{f(x)}{g(x)}$, then: $$ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} $$

A common mnemonic for the Quotient Rule is "low dee high minus high dee low, over the square of what's below." Match each part to the formula $\frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$.

'low' is $g(x)$, 'dee high' is $f'(x)$, 'high' is $f(x)$, and 'dee low' is $g'(x)$. 'The square of what's below' is $[g(x)]^2$. $$ \frac{(\text{low})(\text{dee high}) - (\text{high})(\text{dee low})}{(\text{low})^2} $$

State the Quotient Rule using Leibniz notation for a function $\frac{u}{v}$.

$$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{\frac{du}{dx}v - u\frac{dv}{dx}}{v^2} $$

Let $h(x) = \frac{f(x)}{g(x)}$. If $f(1)=4$, $f'(1)=2$, $g(1)=5$, and $g'(1)=-3$, what is $h'(1)$?

Use the formula $h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$. $$ h'(1) = \frac{(2)(5) - (4)(-3)}{[5]^2} = \frac{10 - (-12)}{25} = \frac{22}{25} $$

Let's find the derivative of $y = \frac{2x-5}{x^3+1}$. Identify the 'high' function $f(x)$ and the 'low' function $g(x)$.

The 'high' function is the numerator: $f(x) = 2x-5$. The 'low' function is the denominator: $g(x) = x^3+1$.

Continuing with $y = \frac{2x-5}{x^3+1}$, what are the derivatives 'dee high' $f'(x)$ and 'dee low' $g'(x)$?

'dee high': $f'(x) = 2$. 'dee low': $g'(x) = 3x^2$.

Assemble the Quotient Rule formula for $y = \frac{2x-5}{x^3+1}$ using the functions and derivatives from the previous steps.

$$ y' = \frac{(2)(x^3+1) - (2x-5)(3x^2)}{(x^3+1)^2} $$

Find the simplified derivative of $y = \frac{2x-5}{x^3+1}$ by expanding the numerator.

$$ y' = \frac{(2x^3+2) - (6x^3-15x^2)}{(x^3+1)^2} = \frac{2x^3+2 - 6x^3+15x^2}{(x^3+1)^2} = \frac{-4x^3+15x^2+2}{(x^3+1)^2} $$

Find the derivative of $y = \frac{7}{x^4}$ using the Quotient Rule.

Let $f(x)=7$ and $g(x)=x^4$. Then $f'(x)=0$ and $g'(x)=4x^3$. $$ y' = \frac{(0)(x^4) - (7)(4x^3)}{(x^4)^2} = \frac{-28x^3}{x^8} = -\frac{28}{x^5} $$

Could you find the derivative of $y = \frac{7}{x^4}$ without using the Quotient Rule?

Yes. It's more efficient to rewrite it as $y = 7x^{-4}$ and use the Power Rule: $y' = 7(-4x^{-5}) = -28x^{-5} = -\frac{28}{x^5}$.

When is it generally more efficient to use the Power Rule instead of the Quotient Rule?

It is more efficient to avoid the Quotient Rule when either the numerator or denominator is a constant. 1. Constant Numerator: For a function like $y = \\frac{c}{g(x)}$, it is often easier to use other rules. For example, $y = \\frac{7}{x^4} = 7x^{-4}$ can be solved with the Power Rule. 2. Constant Denominator: For a function like $y = \\frac{f(x)}{c}$, rewrite it as $y = \\frac{1}{c}f(x)$ and use the Constant Multiple Rule.

Find the derivative of $y = \frac{x^2+3x+2}{x+1}$.

First, try to simplify algebraically. The numerator factors: $y = \frac{(x+1)(x+2)}{x+1} = x+2$ for $x \neq -1$. The derivative is simply $y' = 1$. Always look for simplifications before differentiating!

Find the slope of the tangent line to the curve $y = \frac{x}{x^2+1}$ at the origin $(0,0)$.

1. Find the derivative: $y' = \frac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$. 2. Evaluate the derivative at $x=0$ to find the slope: $m = y'(0) = \frac{1-0^2}{(0^2+1)^2} = 1$.

The Quotient Rule for Differentiation

Confirm Action

Prompt