# What is the derivative of f(x) = (x)/(sqrt(7-3x))?

Jun 7, 2015

We can rewrite the denominator using a property of exponentials that states ${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$:

$f \left(x\right) = \frac{x}{7 - 3 x} ^ \left(\frac{1}{2}\right)$

Also, another law of exponentials states that ${a}^{-} n = \frac{1}{a} ^ n$. Thus,

$f \left(x\right) = x {\left(7 - 3 x\right)}^{- \frac{1}{2}}$

Now, we can use the product rule, where $g \left(x\right) = x$ and $h \left(x\right) = {\left(7 - 3 x\right)}^{- \frac{1}{2}}$, following the rule statement:

$y = g \frac{x}{h} \left(x\right)$ then $y ' = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

Thus, we need to find $g ' \left(x\right)$ and $h ' \left(x\right)$ and, then, proceed to the chain rule.

• $g ' \left(x\right) = 1$

• To find $h ' \left(x\right)$, we need chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$. Renaming $u = 7 - 3 x$, we have $h \left(x\right) = {u}^{- \frac{1}{2}}$, which we can derivate using power rule:

$h ' \left(x\right) = - \frac{1}{2} {u}^{- \frac{3}{2}} \left(2 x\right) = - \frac{\cancel{2} x}{\cancel{2} {\left(7 - 3 x\right)}^{\frac{3}{2}}}$

Now, proceeding to the product rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{7 - 3 x} ^ \left(\frac{1}{2}\right) - \frac{{x}^{2}}{7 - 3 x} ^ \left(\frac{3}{2}\right)$

An exponential law states that ${a}^{n} \cdot {a}^{m} = {a}^{n + m}$, thus, we can sum the expressions having ${\left(7 - 3 x\right)}^{\frac{3}{2}}$ as our l.c.d.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{7 - 3 x + {x}^{2}}{7 - 3 x} ^ \left(\frac{3}{2}\right)}$