# What is the derivative? sqrt( (ax) / (a^2 + x^2)

Oct 4, 2017

$\frac{d}{\mathrm{dx}} \sqrt{\frac{a x}{{a}^{2} + {x}^{2}}} = \frac{1}{2 \sqrt{a x}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$

#### Explanation:

We seek:

$\frac{d}{\mathrm{dx}} \sqrt{\frac{a x}{{a}^{2} + {x}^{2}}}$

We need to apply the chain rule and the quotient rule. For clarity we will perform explicit substitutions for the product rule and the chain rule, with practice this can be done in situ:

Let us use the following variables:

Let { (u,=ax, ), (v,=a^2+x^2, ), (w,=u/v, =(ax) / (a^2 + x^2) ), (y,=sqrt(w),=sqrt((ax) / (a^2 + x^2))) :}

Then let us start some differentiation, we start by differentiating the two parts of the quotient $u$ and $v$ wrt $x$:

$\frac{\mathrm{du}}{\mathrm{dx}} = a$ ..... [A]

$\frac{\mathrm{dv}}{\mathrm{dx}} = 2 x$ ..... [B]

And also, we can differentiate $y$ wrt $w$:

$\frac{\mathrm{dy}}{\mathrm{dw}} = \frac{1}{2} {w}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{w}}$ ..... [C}

So we can apply the quotient rule to find the derivative of $w$ wrt $x$:

$\frac{\mathrm{dw}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$ ........ "The Quotient Rule"

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left({a}^{2} + {x}^{2}\right) \left(a\right) - \left(a x\right) \left(2 x\right)}{{a}^{2} + {x}^{2}} ^ 2 \setminus \setminus \setminus \setminus$ Using [A] and [B]

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{a}^{3} + a {x}^{2} - 2 a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$ ..... [D]

We can use the chain rule to get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dw}} . \frac{\mathrm{dw}}{\mathrm{dx}}$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{w}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2 \setminus \setminus \setminus \setminus$ from [C] & [D]

Now we can replace $w$ with our earlier substitution:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\left(\frac{u}{v}\right)}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$

$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2} . \sqrt{\frac{{a}^{2} + {x}^{2}}{a x}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$

$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{a x}} . {\left({a}^{2} + {x}^{2}\right)}^{\frac{1}{2}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$

$\setminus \setminus \setminus \setminus \setminus = \frac{1}{2 \sqrt{a x}} . \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$

With practice we can just write these derivatives by applying the rules as we go and using the chain rule implicitly

$y = {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{\frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dx}} \left(\frac{a x}{{a}^{2} + {x}^{2}}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{- \frac{1}{2}} \frac{\left({a}^{2} + {x}^{2}\right) \left(a\right) - \left(2 x\right) \left(a x\right)}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(\frac{a x}{{a}^{2} + {x}^{2}}\right)}^{- \frac{1}{2}} \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} {\left(a x\right)}^{- \frac{1}{2}} \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2 {\left(a x\right)}^{\frac{1}{2}}} \frac{{a}^{3} - a {x}^{2}}{{a}^{2} + {x}^{2}} ^ \left(\frac{3}{2}\right)$

Giving us the same result with much lkess work.