The Quotient Rule tells us how to differentiate expressions that are the quotient of 2 simpler expressions. This rule applies to functions where \(f(x)\) and \(g(x)\) are differentiable and \(g \neq 0\).
It can be expressed algebraically as:
\(h'(x) = \left[\cfrac{\textcolor{red}{f(x)}}{\textcolor{blue}{g(x)}}\right]' = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{\textcolor{blue}{g^2(x)}}\)
In short, Quotient Rule takes the derivative of \(f(x)\) multiplied by \(g(x)\), subtracts it by \(f(x)\) multiplied by the derivative of \(g(x)\), and divides it all by \(g(x)\) squared.
Similar to Product Rule, the derivative of a quotient is not the same as the quotient of the derivatives!
Differentiate the expression \(q(x) = \cfrac{6x-5}{x^3 + 4}\).
First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives:
\(\textcolor{red}{f(x) = 6x-5}\)
\(\textcolor{red}{f'(x) = 6}\)
\(\textcolor{blue}{g(x) = x^3 + 4}\)
\(\textcolor{blue}{g'(x) = 3x^2}\)
Next, we can differentiate the expression by using the Quotient Rule formula:
\(q'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{\textcolor{blue}{g(x)}^2}\)
\(q'(x) = \cfrac{(\textcolor{blue}{x^3+4})(\textcolor{red}{6}) - (\textcolor{red}{6x-5})(\textcolor{blue}{3x^2})}{\textcolor{blue}{(x^3+4)^2}}\)
\(q'(x) = \cfrac{6x^3+24 - (18x^3 - 15x^2)}{(x^3+4)^2}\)
Then, we can simplify this expression by collecting like terms:
\(q'(x) = \cfrac{6x^3+24- 18^3 + 15x^2}{(x^3+4)^2}\)
\(q'(x) = \cfrac{-12x^3 + 15x^2 + 24}{(x^3+4)^2}\)
Therefore, we can determine that \(q'(x)\) is \(\boldsymbol{\cfrac{-12x^3 + 15x^2 + 24}{(x^3+4)^2}}\).
First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives:
\(\textcolor{red}{f(x) = x+3}\)
\(\textcolor{red}{f'(x) = 1}\)
\(\textcolor{blue}{g(x) = x^2-1}\)
\(\textcolor{blue}{g'(x) = 2x}\)
Then, we can differentiate the entire expression by using the Quotient Rule formula:
\(p'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{\textcolor{blue}{g(x)}^2}\)
\(p'(x) = \cfrac{(\textcolor{blue}{x^2-1})(\textcolor{red}{1}) - (\textcolor{red}{x+3})(\textcolor{blue}{2x})}{(\textcolor{blue}{x^2-1})^2}\)
\(p'(x) = \cfrac{x^2-1 - (2x^2+ 6x)}{(x^2-1)^2}\)
Finally, we can simplify this expression by collecting like terms:
\(p'(x) = \cfrac{x^2-1 - 2x^2 - 6x}{(x^2-1)^2}\)
\(p'(x) = \cfrac{-x^2 - 6x - 1}{(x^2-1)^2}\)
\(p'(x) = -\cfrac{x^2 + 6x + 1}{(x^2-1)^2}\)
Therefore, we can determine that \(p'(x)\) is \(\boldsymbol{-\cfrac{x^2 + 6x + 1}{(x^2-1)^2}}\).