Question

Differentiate the given function with respect to x, \[y=\sqrt{{{e}^{\sqrt{x}}}}\text{ }\!\!~\!\!\text{ , }\!\!~\!\!\text{ }x>0\]

Answer 1

Hint: We can also write the function $\sqrt{{{e}^{\sqrt{x}}}}$ in the form ${{\text{(}{{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$. As we can see \[\sqrt{{{e}^{\sqrt{x}}}}\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\] involves three functions, which are \[{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}\text{,}{{e}^{\sqrt{x}}}\text{,}\sqrt{x}\text{ }\!\!~\!\!\text{ }\]. We can use the chain rule to solve this question.

Complete step-by-step answer:
The function given in the question is \[y=\sqrt{{{e}^{\sqrt{x}}}}\text{ }\!\!~\!\!\text{ , }\!\!~\!\!\text{ }x>0\]. As we can see, the function has multiple functions. So, for differentiating this type of functions, we can use the chain rule of differentiation.
The chain rule states that:
\[\dfrac{d}{dx}(f(g(x)))=f'(g(x))g'(x)\]
We know that the formula for the derivation of ${{e}^{x}}$and ${{x}^{n}}$ is given by,
 \[\dfrac{d}{dx}({{e}^{x}})={{e}^{x}}\] and \[\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}\]
Let us assume ${{e}^{\sqrt{x}}}$ in the place of $x$ such that ${{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$ reduces into form ${{x}^{n}}$.
Therefore, we can see that we have \[{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}\] as in the above form of derivative.
Now, applying the formula \[\dfrac{d}{dx}({{x}^{n}})\] and chain rule, we get,
\[\therefore \dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}-1}}\dfrac{d}{dx}({{e}^{\sqrt{x}}})\]
\[=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\dfrac{d}{dx}({{e}^{\sqrt{x}}})................(i)\]
Let us assume \[\sqrt{x}\] in the place of \[x\] such that ${{e}^{\sqrt{x}}}$ reduces into form \[({{e}^{x}})\].
Therefore, we can see that we have ${{e}^{\sqrt{x}}}$ as in the above form of derivative.
Applying the formula for \[\dfrac{d}{dx}({{e}^{x}})\] and chain rule we get,
\[\therefore \dfrac{d}{dx}({{e}^{\sqrt{x}}})={{e}^{\sqrt{x}}}\dfrac{d}{dx}\sqrt{x}...........(ii)\]
We can convert $\sqrt{x}$ in the form ${{x}^{\dfrac{1}{2}}}$.
Then, applying the formula for $\dfrac{d}{dx}({{x}^{n}})$ we get,
\[\therefore \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}}\]
\[=\dfrac{1}{2\sqrt{x}}...............(iii)\]
From equation (i), we get:
\[\dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\dfrac{d}{dx}{{(e)}^{\sqrt{x}}}\]
From equation (ii), we get:
\[\dfrac{d}{dx}({{e}^{\sqrt{x}}})={{e}^{\sqrt{x}}}\dfrac{d}{dx}(\sqrt{x})\]
From equation (iii), we get:
$\dfrac{d}{dx}(\sqrt{x})=\dfrac{1}{2\sqrt{x}}$
Substituting the values from equations (i), (ii), (iii) to the question:
$\dfrac{dy}{dx}=\dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}$
From equation (i),
\[\dfrac{d}{dx}{{({{e}^{\sqrt{x}}})}^{\dfrac{1}{2}}}=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}\dfrac{d}{dx}{{(e)}^{\sqrt{x}}}\]
Substituting the value of $\dfrac{d}{dx}({{e}^{\sqrt{x}}})$ from equation (ii),
$=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}{{e}^{\sqrt{x}}}\dfrac{d}{dx}(\sqrt{x})$
Substituting the value of $\dfrac{d}{dx}(\sqrt{x})$ from equation (iii),
$=\dfrac{1}{2\sqrt{{{e}^{\sqrt{x}}}}}{{e}^{\sqrt{x}}}\dfrac{1}{2\sqrt{x}}$
\[=\dfrac{{{e}^{\sqrt{x}}}}{4\sqrt{{{e}^{\sqrt{x}}}x}}\]
Hence, \[\dfrac{dy}{dx}=\dfrac{{{e}^{\sqrt{x}}}}{4\sqrt{{{e}^{\sqrt{x}}}x}}\]

Note: In this question, x>0 is given. If it was x<0, then we would have had imaginary roots. So, in that case, we would have multiple answers and we would be asked to find which of those couldn’t be the value of x. Before applying chain rule, you have to take care that the question involves function inside the given function, only then the chain rule can be applied. If the question involves chain rule and we are comfortable using it, then we can directly differentiate each function separately and write the answer. It will take less time, for example:
Differentiate ${{e}^{{{x}^{2}}}}$
$\dfrac{d}{dx}({{e}^{{{x}^{2}}}})$
Here, we have two functions ${{e}^{{{x}^{2}}}},{{x}^{2}}$. So, we can differentiate these two functions separately and simply write the answer as ${{e}^{{{x}^{2}}}}2x$.