# How do you differentiate f(x)=x/(2^sqrt(x-3)) using the chain rule?

Nov 1, 2016

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

#### Explanation:

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

Use quotient rule and chain rule

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

$f ' = 1 ,$ $g ' = {2}^{{\left(x - 3\right)}^{\frac{1}{2}}} \ln 2 \cdot \frac{1}{2} {\left(x - 3\right)}^{- \frac{1}{2}} \cdot 1$

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