Question

Find from first principle the derivative of f(x)=root of X with respect to x

Answer 1

Given:

$f(x)=\sqrt{x}$

Required:

To find the derivative of the given function by using the first principle.

Explanation:

To find the derivative by the first principle we will use the limit method.

$\begin{gathered} f^{\prime}(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\ f^{\prime}(x)=\operatorname{\lim}_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \end{gathered}$

Rationalise the denominator

$f^{\prime}(x)=\operatorname{\lim}_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$

Use the formula:

$(a+b)(a-b)=a^2-b^2$$\begin{gathered} f^{\prime}(x)=\lim_{h\to0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\ f^{\prime}(x)=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ f^{\prime}(x)=\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt{x)}} \end{gathered}$

Now apply the given limit

$\begin{gathered} f^{\prime}(x)=\frac{1}{(\sqrt{x+0}-\sqrt{x})} \\ f^{\prime}(x)=\frac{1}{2\sqrt{x}} \end{gathered}$

Final answer:

Thus the derivative of the given function is

$f^{\prime}(x)=\frac{1}{2\sqrt{x}}$