Question

How do I find the derivative of $f\left( x \right)=\sqrt{x}$ using first principles?

Answer 1

Hint: Here we have to find the derivative of a given function using first principles. Firstly we will write the first principle formula which is ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ and substitute the value of the function given in it. Then we will simplify our value inside the limit sign by rationalizing the numerator method. Finally we will substitute the limit in the value and get our desired answer.

Complete step-by-step answer:
We have to find the derivative of below function using first principles:
$f\left( x \right)=\sqrt{x}$……$\left( 1 \right)$
Now as we know from the definition of first principle that the derivative of the function is calculated by below formula:
 ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Now from equation (1) we get $f\left( x+h \right)=\sqrt{x+h}$ substitute the values above we get,
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$…..$\left( 2 \right)$
If we substitute the limit we will get the value as:
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x+0}-\sqrt{x}}{0}$
$\Rightarrow {f}'\left( x \right)=\dfrac{0}{0}$
As we are getting indeterminate form we will simplify the value further before substituting the limit in it.
Now using rationalizing formula multiply and divide the equation (2) by $\sqrt{x+h}+\sqrt{x}$ as follows:
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\times \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$
Using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ above where $a=\sqrt{x+h},b=\sqrt{x}$ we get,
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\left( \sqrt{x+h} \right)}^{2}}-{{\left( \sqrt{x} \right)}^{2}}}{h\left( \sqrt{x+h}+\sqrt{x} \right)}$
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{x+h-x}{h\left( \sqrt{x+h}+\sqrt{x} \right)}$
Simplifying further we get,
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{h}{h\left( \sqrt{x+h}+\sqrt{x} \right)}$
Cancel out the common term from numerator and denominator as follows:
$\Rightarrow {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{\left( \sqrt{x+h}+\sqrt{x} \right)}$
Now put the limit,
$\Rightarrow {f}'\left( x \right)=\dfrac{1}{\left( \sqrt{x+0}+\sqrt{x} \right)}$
$\Rightarrow {f}'\left( x \right)=\dfrac{1}{2\sqrt{x}}$
So we get the derivative as $\dfrac{1}{2\sqrt{x}}$
Hence the derivative of $f\left( x \right)=\sqrt{x}$ using first principles is $\dfrac{1}{2\sqrt{x}}$ .
So, the correct answer is “$\dfrac{1}{2\sqrt{x}}$”.

Note: The first principle of the derivative states that if $f\left( x \right)$ is a real function in its domain then if a function defined as ${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ exist it is known as the derivative of that function. Rationalizing is used in order to get rid of all the radicals which in this case are the square root values. We have multiplied and divided the same value present in the numerator with the middle sign changed as it will give us a perfect square value which can be easily simplified.