### 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)}#.

According to the

*quotient rule for differentiation*we have\[\begin{array}{rcl}\dfrac{\dd}{\dd x}\left(\dfrac{f}{g}(x)\right) &=& \dfrac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{g(x)^2}\\ &=& \dfrac{(2x+9)\cdot (4\cdot x+2)-4\cdot (x^2+9\cdot x+1)}{(4\cdot x+2)^2}\\ & =& \displaystyle {{2\cdot x^2+2\cdot x+7}\over{8\cdot x^2+8\cdot x+2}} \end{array}\]

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.