Question

Find the slope of the tangent to the curve $y=2\sqrt{x}$ at the point $(1,2)$.

Answer 1

$y=2\sqrt{x}$
$\frac{dy}{dx}=2\left(\frac{1}{\left(2\surd x\right)}\right)=\frac{1}{\sqrt{x}}$
$\frac{dy}{dx}$ is called slope of the tangent to the curve $y=2\sqrt{x}$ at the point $(x,y)$. For the point $x=1$ and $y=2$. Putting $x=1$ and $y=2$ in the equation $\frac{dy}{dx}=\frac{1}{\sqrt{x}}$ , we get $\frac{dy}{dx}=\frac{1}{\sqrt{1}}$. Hence slope is 1.