### 2 Differentiating a quotient

In this Section we consider functions of the form $y=\frac{f\left(x\right)}{g\left(x\right)}$ . To find the derivative of such a function we make use of the following Key Point:

##### Key Point 10

Quotient Rule

If $y=\frac{f\left(x\right)}{g\left(x\right)}$ then $\frac{dy}{dx}=\frac{g\left(x\right)\frac{df}{dx}-\frac{dg}{dx}f\left(x\right)}{{\left[g\left(x\right)\right]}^{2}}$ $\left(\text{or}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{y}^{\prime }=\frac{g{f}^{\prime }-{g}^{\prime }f}{{g}^{2}}\right)$

If $y=\frac{u}{v}$  then $\frac{dy}{dx}=\frac{v\frac{du}{dx}-\frac{dv}{dx}u}{{v}^{2}}$ $\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\left(\text{or}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{y}^{\prime }=\frac{v{u}^{\prime }-{v}^{\prime }u}{{v}^{2}}\right)$

These are two equivalent versions of the quotient rule .

##### Example 10

Find the derivative of $y=\frac{lnx}{x}$

##### Solution

Here $f\left(x\right)=lnx$ and $g\left(x\right)=x$

$\therefore \phantom{\rule{2em}{0ex}}\frac{df}{dx}=\frac{1}{x}$ and $\frac{dg}{dx}=1$

Hence $\frac{dy}{dx}=\frac{x\left(\frac{1}{x}\right)-1\left(lnx\right)}{{\left[x\right]}^{2}}=\frac{1-lnx}{{x}^{2}}$

Obtain the derivative of $y=\frac{sinx}{{x}^{2}}$

1. using the formula for differentiating a product and
2. using the formula for differentiating a quotient.
1. Write $y={x}^{-2}sinx$ then use the product rule to find $\frac{dy}{dx}$ :

$y={x}^{-2}sinx\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}\frac{dy}{dx}=\left(-2{x}^{-3}\right)sinx+{x}^{-2}cosx$ $\therefore \phantom{\rule{2em}{0ex}}\frac{dy}{dx}=\frac{-2sinx+xcosx}{{x}^{3}}$

2. Now use the quotient rule instead to find $\frac{dy}{dx}$ :

$y=\frac{sinx}{{x}^{2}}\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}\frac{dy}{dx}=\frac{{x}^{2}\left(cosx\right)-\left(2x\right)sinx}{{\left({x}^{2}\right)}^{2}}=\frac{xcosx-2sinx}{{x}^{3}}$

##### Exercise

Find the derivatives of the following:

1. $\left(2{x}^{3}-4{x}^{2}\right)\left(3{x}^{5}+{x}^{2}\right)$
2. $\frac{2{x}^{3}+4}{{x}^{2}-4x+1}$
3. $\frac{{x}^{2}+2x+1}{{x}^{2}-2x+1}$
4. $\left({x}^{2}+3\right)\left(2x-5\right)\left(3x+2\right)$
5. $\frac{\left(2x+1\right)\left(3x-1\right)}{x+5}$
6. $\left(lnx\right)sinx$
7. $\left(lnx\right)∕sinx$
8. ${e}^{x}∕{x}^{2}$
9. $\frac{{e}^{x}sinx}{cos2x}$
1. $48{x}^{7}-84{x}^{6}+10{x}^{4}-16{x}^{3}$
2. $\frac{2{x}^{4}-16{x}^{3}+6{x}^{2}-8x+16}{{\left({x}^{2}-4x+1\right)}^{2}}$
3. $-\frac{4\left(x+1\right)}{{\left(x-1\right)}^{3}}$
4. $24{x}^{3}-33{x}^{2}+16x-33$
5. $\frac{6\left({x}^{2}+10x+1\right)}{{\left(x+5\right)}^{2}}$
6. $\frac{1}{x}sinx+\left(lnx\right)cosx$
7. $\frac{sinx\left(\frac{1}{x}\right)-\left(lnx\right)cosx}{{sin}^{2}x}=\text{cosec}x\left(\frac{1}{x}\phantom{\rule{0.3em}{0ex}}-cotxlnx\right)$
8. $\frac{{x}^{2}{e}^{x}-2x{e}^{x}}{{x}^{4}}=\left({x}^{-2}-2{x}^{-3}\right){e}^{x}$
9. $\frac{cos2x\left({e}^{x}sinx+{e}^{x}cosx\right)+2sin2x{e}^{x}sinx}{{cos}^{2}2x}$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}={e}^{x}\left[\left(sinx+cosx\right)sec2x+2sinxsin2x{sec}^{2}2x\right]$