User:Nsh/Chain-rule question

The original problem set is: $f(x)={\frac {x}{\sqrt {3-x}}}$

There are two techniques of differentiation required to solve the problem: Chain_rule and Product_rule
The chain rule because we have a funtion of a function: $f(x)={\frac {x}{\sqrt {g(x)}}}$ where $g(x)=3-x$

And the product rule because we are effectivly differentiating a product: $f(x)=xg(x)^{-{\frac {1}{2}}}$

By the chain rule ( ${\frac {d}{dx}}{f(g(x))}={\frac {d}{dx}}g{\frac {d}{dx}}f$ ) we can use the substitution $u=3-x$
$du/dx=-1$
f`(x) = ${\frac {du}{dx}}x(-{\frac {1}{2}})x^{-{\frac {3}{2}}}+{\frac {x}{\sqrt {3-x}}}={\frac {1}{2}}x^{-{\frac {1}{2}}}+{\frac {x}{\sqrt {3-x}}}$ (Two terms because of Product_rule)