# How do you find the Slope of the curve y=sqrt(x) at the point where x=4?

Aug 26, 2014

The slope of the curve $y = f \left(x\right)$ at $x = a$ is $f ' \left(a\right)$.
Let us look at the posted problem.

By rewriting the square-root as 1/2-power,
$f \left(x\right) = \sqrt{x} = {x}^{\frac{1}{2}}$

By the power rule,
$f ' \left(x\right) = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$

By plugging in $x = 4$,
Slope: $m = f ' \left(4\right) = \frac{1}{2 \sqrt{4}} = \frac{1}{4}$

Hence, the slope is $\frac{1}{4}$.

I hope that this helps.