Question

How do you differentiate $f(x) = \left( {{x^{\dfrac{1}{2}}}} \right)\left( {\log x} \right)$

Answer 1

Hint: We are given the function and are asked to differentiate it. We will use the formula for product rule to solve this problem. Using this formula, we will expand and then find the derivative of each sub function separately. After that, we will substitute it back and solve it.

Formula used:
Product rule of differentiation:
$\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = g\left( x \right)f'\left( x \right) + f\left( x \right)g'\left( x \right)$

Complete step by step answer:
The given question is , we need to differentiate it.
We will have to use the product rule to get this differentiated
The formula of product rule is $\dfrac{d}{{dx}}f\left( x \right)g\left( x \right) = g\left( x \right)f'\left( x \right) + f\left( x \right)g'\left( x \right)$
$ \Rightarrow f\left( x \right) = \left( {{x^{\dfrac{1}{2}}}} \right)\left( {\log x} \right)$
Let us differentiate,
$\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)\left( {\log x} \right)$
Using this product rule,
$\dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)} \right]\left( {\log x} \right) + \left[ {\dfrac{d}{{dx}}\left( {\log x} \right)} \right]\left( {{x^{\dfrac{1}{2}}}} \right)$
Now, we will find the derivatives separately and then substitute it back.
$ \Rightarrow \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right) = \dfrac{1}{2}{x^{\left( {\dfrac{1}{2} - 1} \right)}}$
Solving this,
$ \Rightarrow \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right) = \dfrac{1}{2}{x^{\left( {\dfrac{{ - 1}}{2}} \right)}}$
Derivation of another term,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$
Now, substituting both the values in the problem, we get,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right)} \right]\left( {\log x} \right) + \left[ {\dfrac{d}{{dx}}\left( {\log x} \right)} \right]\left( {{x^{\dfrac{1}{2}}}} \right)$
On rewriting the term and we get,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{1}{2}{x^{\dfrac{{ - 1}}{2}}}} \right]\left( {\log x} \right) + \left[ {\dfrac{1}{x}} \right]\left( {{x^{\dfrac{1}{2}}}} \right)$
Further simplify the term and we get,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{1}{{2{x^{ - 2}}}}} \right]\left( {\log x} \right) + \left[ {\dfrac{{{x^{ - 2}}}}{x}} \right]$
On rewriting the term and we get,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{1}{{2\sqrt x }}} \right]\left( {\log x} \right) + \left[ {\dfrac{{\sqrt x }}{x}} \right]$
Then we get
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{1}{{2\sqrt x }}} \right]\left( {\log x} \right) + \left[ {\dfrac{1}{{\sqrt x }}} \right]$

Therefore, the derivative of the given function is$\dfrac{d}{{dx}}f\left( x \right) = \left[ {\dfrac{1}{{2\sqrt x }}} \right]\left( {\log x} \right) + \left[ {\dfrac{1}{{\sqrt x }}} \right]$

Note: Derivatives are also known as the fundamental tool of calculus. A derivative of any function in real variables measures the sensitivity to change of the function value with respect to a change in its argument.
The derivative also tells us the slope of a function at any point.
There are rules to solve or to do derivation.
For any function and, we can use the following formulas to find derivatives in specific cases respectively.

Common functions	Function	Derivative
Sum Rule	$f + g$	$f' + g'$
Difference Rule	$f - g$	$f' - g'$
Product Rule	$fg$	$fg' + f'g$
Quotient Rule	$\dfrac{f}{g}$	$\dfrac{{f'g + g'f}}{{{g^2}}}$

Make sure to remember all these rules because these are very important and most frequently asked or used rules.