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

Aug 11, 2017

We are going to use the simple rule of the derivative of a power function $\mathrm{df} {\left(x\right)}^{n} / \mathrm{dx} = n {f}^{n - 1} \mathrm{df} \frac{x}{\mathrm{dx}}$ to find, posing

$g \left(x\right) = \setminus \sqrt{\setminus \frac{3 {x}^{2}}{2 x - 3}} = {\left[\setminus \frac{3 {x}^{2}}{2 x - 3}\right]}^{\setminus} \frac{1}{2}$
and
$h \left(x\right) = {e}^{g} \left(x\right)$
that
$\setminus \frac{\mathrm{dh}}{\mathrm{dx}} = \setminus \frac{\mathrm{dg}}{\mathrm{dx}} h$
$\setminus \frac{\mathrm{dh}}{\mathrm{dx}} = \frac{1}{2} {\left[\setminus \frac{3 {x}^{2}}{2 x - 3}\right]}^{\setminus} \frac{- 1}{2} \setminus \frac{\left(2 x - 3\right) 6 x - 3 {x}^{2} \left(2\right)}{{\left(2 x - 3\right)}^{2}}$
where I have used the chain rule for the ratio of two functions
$\setminus \frac{\mathrm{dh}}{\mathrm{dx}} = \setminus \frac{6 {x}^{2} - 9 x - 3 {x}^{2}}{\left[{\left(2 x - 3\right)}^{2}\right]} {\left[\setminus \frac{3 {x}^{2}}{2 x - 3}\right]}^{\setminus} \frac{- 1}{2}$
$\setminus \frac{\mathrm{dh}}{\mathrm{dx}} = \setminus \frac{3 {x}^{2} - 9 x}{{\left[3 {x}^{2}\right]}^{\setminus} \frac{1}{2}} {\left(2 x - 3\right)}^{\setminus} \frac{- 5}{2}$
$\setminus \frac{\mathrm{dh}}{\mathrm{dx}} = \sqrt{3} \left(x - 3\right) {\left(2 x - 3\right)}^{\setminus} \frac{- 5}{2}$