The quotient rule for differentiation

Rules of differentiation: Rules of computation for the derivative

The quotient rule for differentiation

Let $f$ and $g$ be functions. The quotient function $\dfrac{f}{g}$ is the function that assigns the value $\dfrac{f(x)}{g(x)}$ to $x$ .

For example, if $f(x)=x+1$ and $g(x)=x^3+1$ , then the quotient function $\frac{f}{g}$ has function rule

$\frac{f}{g}(x) = \frac{f(x)}{g(x)}=\frac{x+1}{x^3+1}\tiny.$

The following rule determines the derivative of a quotient function. Recall that $g^2$ for a function $g$ stands for $g\cdot g$ .

Quotient rule for differentiation

Let $f$ and $g$ be differentiable functions. The derivative of $\dfrac{f}{g}$ is $\dfrac{f'\cdot g-g'\cdot f}{g^2}$ , so $\left(\dfrac{f}{g}\right)'(x) = \dfrac{f'(x)\cdot g(x)-f(x) \cdot g'(x)}{g(x)^2}\tiny.$

Let $h(x)=\dfrac{f(x)}{g(x)}$ . We want to prove that $h'(x)=\dfrac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}$ .

We will use the definition of $h$ in the form $f(x)=g(x) \cdot h(x)$ . The product rule then says that $f'(x)=g'(x) \cdot h(x) + g(x) \cdot h'(x)$ . This means that: $h'(x)=\frac{f'(x)-g'(x)\cdot h(x)}{g(x)}$ .

Since $h(x)=\dfrac{f(x)}{g(x)}$ , we have $h'(x)=\frac{f'(x) \cdot g(x)-g'(x)\cdot f(x)}{g^2(x)}$ .

So $h'(x)=\left(\dfrac{f}{g}\right)'(x)=\frac{f'(x) \cdot g(x)-f(x)\cdot g'(x)}{g^2(x)}$ .

Determine the derivative of the function $\dfrac{f}{g}$ if $f(x) = x^2+9\cdot x-2$ and $g(x) = 3\cdot x-4$ .

${{3\cdot x^2-8\cdot x-30}\over{9\cdot x^2-24\cdot x+16}}$

According to the quotient rule for differentiation we have $\begin{array}{rcl}\dfrac{\dd}{\dd x}\left(\dfrac{f}{g}(x)\right) &=& \dfrac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}\\ &=& \dfrac{(2x+9)\cdot (3\cdot x-4)-3\cdot (x^2+9\cdot x-2)}{(3\cdot x-4)^2}\\ & =& \displaystyle {{3\cdot x^2-8\cdot x-30}\over{9\cdot x^2-24\cdot x+16}} \end{array}$

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