Differentiating $y=x^{-3/2}$ I'd like to differentiate $y=x^{\large\frac{-3}{2}}$. So, 
$$y + dy = (x + dx)^{\large\frac{-3}{2}} = x^{\large\frac{-3}{2}}\Big(1 + \frac{dx}{x}\Big)^{\large\frac{-3}{2}}$$
is at least a start. How do I calculate the parentheses?

Try squaring both sides and then eliminating less significant ${dy}^2$, ${dx}^2$, $dxdy$ and ${dx}^3$ etc. kind of terms.
Then you will get it.$$(y+dy)^2 = \frac{1}{(x+dx)^3}$$  $$\implies y^2+{dy}^2+2y\cdot dy = \frac{1}{x^3+{dx}^3+3x^2\cdot dx+3x\cdot {dx}^2}$$
Eliminating less significant higher degrees of $dx$ and $dy$
$$(y^2+2y\cdot dy)(x^3+3x^2\cdot dx) = 1$$$$\implies x^3y^2+2x^3y\cdot dy+3x^2y^2\cdot dx = 1$$
Using $y = x^{-3/2}$ here, it becomes
$$ 1+2x^{3/2}\cdot dy+3x^{-1}\cdot dx=1$$
$$\implies \frac{dy}{dx}=-\frac{3}{2}{x^{-5/2}}$$
Hope that helps.
