Answer to Question #121874 in Calculus for desmond

Question

Use the method of Differentiation from the first principle to prove that the
the derivative of the square root of x is given below
d/dx x^1/2=1/2x^1/2

Answer 1

This is the formula for finding the derivative from the first principle:

$\frac{dy}{dx}=\lim_{\Delta x \to 0} \frac {f(x+\Delta x)-f(x)}{\Delta x}.$

So we have

$\frac{d}{dx}(x^{\frac{1}{2}})=\frac{d}{dx}(\sqrt x)=\lim_{\Delta x \to 0} \frac {\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}=$

$=\lim_{\Delta x \to 0} \frac {\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x} \cdot \frac {\sqrt{x+\Delta x}+\sqrt{x}}{\sqrt{x+\Delta x}+\sqrt{x}}=$

$=\lim_{\Delta x \to 0} \frac {(\sqrt{x+\Delta x})^2-(\sqrt{x})^2}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})} =\lim_{\Delta x \to 0} \frac {x+\Delta x-x}{\Delta x(\sqrt{x+\Delta x}+\sqrt{x})}=$

$=\lim_{\Delta x \to 0} \frac {1}{\sqrt{x+\Delta x}+\sqrt{x}}= \frac {1}{\sqrt{x}+\sqrt{x}}= \frac {1}{2\sqrt{x}}=\frac {1}{2x^{\frac{1}{2}}}.$