For $y=\sqrt{x}$, with $x0$: rewrite $y$ using a fractional index; find \(\dfrac{\mat

Question

For $y=\sqrt{x}$, with $x>0$: rewrite $y$ using a fractional index; find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$; find the gradient at $x=9$.

Accepted Answer

(a) $y=x^{\frac12}$; $y'=\frac{1}{(2\sqrt{x})}$; (c) Gradient $\frac16$.

Differentiation