# How do you find (d^2y)/(dx^2) given y=x^(9/4)?

The answer is $\frac{45}{16} \cdot {x}^{\frac{1}{4}}$
We use $\left({x}^{n}\right) ' = n {x}^{n - 1}$
$y = {x}^{\frac{9}{4}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{9}{4} \cdot {x}^{\frac{9}{4} - 1} = \frac{9}{4} {x}^{\frac{5}{4}}$
$\frac{{d}^{2} x}{\mathrm{dy}} ^ 2 = \frac{9}{4} \cdot \frac{5}{4} {x}^{\frac{5}{4} - 1} = \frac{45}{16} \cdot {x}^{\frac{1}{4}}$