For $$y=\frac1{x^3},\qquad x\ne0,$$ rewrite $y$ using a negative index; find \(\dfrac{\mathrm{d}y}

Question

For $$y=\frac1{x^3},\qquad x\ne0,$$ rewrite $y$ using a negative index; find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$; find the gradient at $x=2$.

Accepted Answer

(a) $y=x^{-3}$; (b) $y'=-3x^{-4}=-\frac{3}{x^4}$; (c) Gradient $-\frac{3}{16}$.

Differentiation