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
 ${\displaystyle \frac{d}{dx}}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\times g^{\prime }\left(x\right)$

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
 ${\displaystyle \frac{d}{dx}}f\left(g\left(x\right)\right)=f^{\prime }\left(g\left(x\right)\right)\times g^{\prime }\left(x\right)$
So we take our function and differentiate it with respect to $x$ and have
 ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}\left(e^{\sqrt{x}}\right)$
Now we remove the square root of $x$ and take a power of half to solve further
$${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{d}{dx}}\left(e^{x^{{}^{{\displaystyle \frac{1}{2}}}}}\right)=e^{x^{{}^{{\displaystyle \frac{1}{2}}}}}\times {\displaystyle \frac{1}{2}}{x}^{\left({\displaystyle \frac{1}{2}}-1\right)}=e^{\sqrt{x}}\times {\displaystyle \frac{1}{2}}{x}^{-{\displaystyle \frac{1}{2}}}=\left({\displaystyle \frac{e^{\sqrt{x}}}{2}}\right)x^{-{\displaystyle \frac{1}{2}}}$$
The formula used here is given below
 ${\displaystyle \frac{d}{dx}}\left(x^n\right)=n{x}^{n-1}$
We have got the derivative as $$\left({\displaystyle \frac{e^{\sqrt{x}}}{2}}\right)x^{-{\displaystyle \frac{1}{2}}}$$ 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
 ${\displaystyle \frac{dy}{dx}}=\left({\displaystyle \frac{e^{\sqrt{x}}}{2}}\right)x^{-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{e^{\sqrt{x}}}{2x^{{\displaystyle \frac{1}{2}}}}}={\displaystyle \frac{e^{\sqrt{x}}}{2\sqrt{x}}}$
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
 ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{e^{\sqrt{x}}}{2\sqrt{x}}}\times {\displaystyle \frac{\sqrt{x}}{\sqrt{x}}}={\displaystyle \frac{e^{\sqrt{x}}\sqrt{x}}{2x}}\left(\because \sqrt{a}\times \sqrt{a}=a\right)$
In the denominator part both the square root expressions get cancelled out giving the derivative.

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

Note: We could also use the direct formula for calculating the derivative of $\sqrt{x}$ with respect to $x$ given by ${\displaystyle \frac{d\sqrt{x}}{dx}}={\displaystyle \frac{1}{2\sqrt{x}}}$ 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.