Question

Find the derivative of the function: $y = {e^{\sqrt x }}$

Answer 1

Hint : We find the derivative of the given composite function with respect to $x$ and use chain rule which says that the derivative of a composite function will be equal to the derivative of the outside function with respect to inside times the derivative of inside function, mathematically it can be seen as
$\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$

Complete step-by-step answer :
Firstly we write down the function given in the question
$y = {e^{\sqrt x }}$
As we can see the function is a composite function because of the two functions together as exponential function and a square root of $x$ together. So while evaluating derivatives of such composite function we must apply chain rule to solve it i.e. the derivative of a composite function will be equal to the derivative of the outside function with respect to inside times the derivative of inside function, mathematically it can be seen as
$\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$
So we take our function and differentiate it with respect to $x$ and have
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{\sqrt x }}} \right)$
Now we remove the square root of $x$ and take a power of half to solve further
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{{x^{^{\dfrac{1}{2}}}}}}} \right) = {e^{{x^{^{\dfrac{1}{2}}}}}} \times \dfrac{1}{2}{x^{\left( {\dfrac{1}{2} - 1} \right)}} = {e^{\sqrt x }} \times \dfrac{1}{2}{x^{ - \dfrac{1}{2}}} = \left( {\dfrac{{{e^{\sqrt x }}}}{2}} \right){x^{ - \dfrac{1}{2}}}$
The formula used here is given below
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
We have got the derivative as $\left( {\dfrac{{{e^{\sqrt x }}}}{2}} \right){x^{ - \dfrac{1}{2}}}$ which has a negative exponent so to change it into a positive exponent we shift it in the denominator like this and simplify it further
$\dfrac{{dy}}{{dx}} = \left( {\dfrac{{{e^{\sqrt x }}}}{2}} \right){x^{ - \dfrac{1}{2}}} = \dfrac{{{e^{\sqrt x }}}}{{2{x^{\dfrac{1}{2}}}}} = \dfrac{{{e^{\sqrt x }}}}{{2\sqrt x }}$
This can further be simplified because we have a square root in the denominator part. To remove that we rationalize the fraction and multiply both numerator and denominator by $\sqrt x$ like this
$\dfrac{{dy}}{{dx}} = \dfrac{{{e^{\sqrt x }}}}{{2\sqrt x }} \times \dfrac{{\sqrt x }}{{\sqrt x }} = \dfrac{{{e^{\sqrt x }}\sqrt x }}{{2x}}\left( {\because \sqrt a \times \sqrt a = a} \right)$
In the denominator part both the square root expressions get cancelled out giving the derivative.

So, the correct answer is “$\dfrac{{{e^{\sqrt x }}\sqrt x }}{{2x}}$”.

Note : We could also use the direct formula for calculating the derivative of $\sqrt x$ with respect to $x$ given by $\dfrac{{d\sqrt x }}{{dx}} = \dfrac{1}{{2\sqrt x }}$ directly in the question to get the answer quickly. Remembering such formulae or results directly can save us a lot of time while evaluating derivatives.