Differentiate the function; f(x)=1/((5-2x^3)^2)

We know from the properties of basic indices that a^-x=1/a^x, so 1/((5-2x³)²=(5-2x³)^-2 where in this case, a=5-2x³and x=2. Then the function is differentiable by the chain rule. As dy/dx=dy/duXdu/dx, we let u=5-2x³, and by the principles of differentiation, du/dx=-6x². If f(x)=y=(5-2x³)^-2, we have that y=u^-2, hence dy/du=-2u^-3. therefore by the chain rule, dy/dx=dy/duXdu/dx=-2u^-3X-6x²=12x²u^-3=12x²(5-2x³)^-3=12x²/(5-2x³)³.
So when f(x)=1/(5-2x³)², f'(x)=12x²/(5-2x³)³.