# Question 022c2

Jun 6, 2015

The following derivative can be solved by chain rule.

Let us substitute $\frac{x - 3}{x} ^ 2 = {u}^{2}$
$\therefore \sqrt{\frac{x - 3}{x} ^ 2} = u$

Differentiating the original function $f$ w.r.t. $u$,
$\frac{\mathrm{df}}{\mathrm{du}} = \frac{d}{\mathrm{du}} \left(u\right)$
$\implies \frac{\mathrm{df}}{\mathrm{du}} = 1$

Now, we are required to determine $\frac{\mathrm{df}}{\mathrm{dx}}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Differentiating w.r.t. $x$ on both sides of the expression $\frac{x - 3}{x} ^ 2 = {u}^{2}$ we get ( applying quotient rule ),

$\frac{{x}^{2} - \left(x - 3\right) \cdot 2 x}{x} ^ 4 = \frac{d}{\mathrm{dx}} \left({u}^{2}\right)$
$\implies \frac{6 - {x}^{2}}{x} ^ 3 = \frac{\mathrm{du}}{\mathrm{dx}} .2 u$
=>(du)/dx = (6-x^2)/(x^3.(2u)
=>(du)/dx = (6-x^2)/(x^3.(2.sqrt ( (x-3)/x^2) )

:. (df)/dx = 1.(6-x^2)/(x^3.(2.sqrt ( (x-3)/x^2) )# [$\frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$]