How do you differentiate \[y={{x}^{\sqrt{x}}}\]?

Hint: This question is from the topic of calculus. In solving this question, we will first put log base e (that is \[\ln \]) to the both sides of the equation. After that, we will differentiate both sides of the equation. We will use the formula \[d\left( \ln x \right)=\left( \dfrac{1}{x} \right)dx\] and \[d\left( {{x}^{n}} \right)=\left( n{{x}^{n-1}} \right)dx\] for the further differentiation. After that, we will solve the further question and find the value of \[\dfrac{dy}{dx}\] that will be our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to differentiate \[y={{x}^{\sqrt{x}}}\]. Or, we can say that we have to find the value of \[\dfrac{dy}{dx}\] using the equation \[y={{x}^{\sqrt{x}}}\].
The term which we have to solve is
\[y={{x}^{\sqrt{x}}}\]
Now, putting \[\ln \](that is log base ‘e’ or we can say \[{{\log }_{e}}\]) to the both side of the equation, we get
\[\Rightarrow \ln y=\ln {{x}^{\sqrt{x}}}\]
Now, using the formula of logarithms that is \[\ln {{x}^{a}}=a\ln x\], we can write the above equation as
\[\Rightarrow \ln y=\sqrt{x}\ln x\]
Now, differentiating both side of equation, we can write
\[\Rightarrow d\left( \ln y \right)=d\left( \sqrt{x}\ln x \right)\]
Now, using the formula of product rule that is \[d\left( u.v \right)=vd\left( u \right)-ud\left( v \right)\], we can write the above equation as
\[\Rightarrow d\left( \ln y \right)=\ln x\centerdot d\left( \sqrt{x} \right)+\sqrt{x}\centerdot d\left( \ln x \right)\]
\[\Rightarrow d\left( \ln y \right)=\ln x\centerdot d\left( {{x}^{\dfrac{1}{2}}} \right)+\sqrt{x}\centerdot d\left( \ln x \right)\]
Now, using the formula of differentiation that is \[d\left( {{x}^{n}} \right)=\left( n{{x}^{n-1}} \right)dx\], we can write
\[\Rightarrow d\left( \ln y \right)=\ln x\centerdot \left( \dfrac{1}{2}{{x}^{\dfrac{1}{2}-1}} \right)dx+\sqrt{x}\centerdot d\left( \ln x \right)\]
\[\Rightarrow d\left( \ln y \right)=\ln x\centerdot \left( \dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \right)dx+\sqrt{x}\centerdot d\left( \ln x \right)\]
Now, using the formula of differentiation that is \[d\left( \ln x \right)=\left( \dfrac{1}{x} \right)dx\], we can write the above equation as
\[\Rightarrow \left( \dfrac{1}{y} \right)dy=\ln x\centerdot \left( \dfrac{1}{2}{{x}^{-\dfrac{1}{2}}} \right)dx+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)dx\]
The above equation can also be written as
\[\Rightarrow \left( \dfrac{1}{y} \right)dy=\ln x\centerdot \left( \dfrac{1}{2\left( {{x}^{\dfrac{1}{2}}} \right)} \right)dx+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)dx\]
\[\Rightarrow \left( \dfrac{1}{y} \right)dy=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)dx+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)dx\]
Now dividing ‘dx’ to both side of the equation, we can write
\[\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)+\sqrt{x}\centerdot \left( \dfrac{1}{x} \right)\]
The above equation can also be written as
\[\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)+\sqrt{x}\centerdot \left( \dfrac{1}{\sqrt{x}\centerdot \sqrt{x}} \right)\]
\[\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\ln x\centerdot \left( \dfrac{1}{2\sqrt{x}} \right)+\left( \dfrac{1}{\sqrt{x}} \right)\]
The above equation can also be written as
\[\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\dfrac{\ln x}{2}\left( \dfrac{1}{\sqrt{x}} \right)+\left( \dfrac{1}{\sqrt{x}} \right)\]
Taking \[\dfrac{1}{\sqrt{x}}\] as common in the right side of the equation, we can write
\[\Rightarrow \left( \dfrac{1}{y} \right)\dfrac{dy}{dx}=\left( \dfrac{1}{\sqrt{x}} \right)\left( \dfrac{\ln x}{2}+1 \right)\]
\[\Rightarrow \dfrac{dy}{dx}=y\left( \dfrac{1}{\sqrt{x}} \right)\left( \dfrac{\ln x}{2}+1 \right)\]
Now, putting the value of y in the above equation, we can write
\[\Rightarrow \dfrac{dy}{dx}={{x}^{\sqrt{x}}}\left( \dfrac{1}{\sqrt{x}} \right)\left( \dfrac{\ln x}{2}+1 \right)\]
\[\Rightarrow \dfrac{dy}{dx}={{x}^{\sqrt{x}}}\left( {{x}^{-\dfrac{1}{2}}} \right)\left( \dfrac{\ln x}{2}+1 \right)\]
Now, using the formula \[{{x}^{a}}{{x}^{b}}={{x}^{a+b}}\], we can write
\[\Rightarrow \dfrac{dy}{dx}={{x}^{\sqrt{x}-\dfrac{1}{2}}}\left( \dfrac{\ln x}{2}+1 \right)\]
Or, we can write the above equation as
\[\Rightarrow \dfrac{dy}{dx}={{x}^{-\dfrac{1}{2}+\sqrt{x}}}\left( \dfrac{\ln x+2}{2} \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}+\sqrt{x}}}\left( \ln x+2 \right)\]
Hence, we have done differentiation now. The differentiation is \[\dfrac{dy}{dx}=\dfrac{1}{2}{{x}^{-\dfrac{1}{2}+\sqrt{x}}}\left( \ln x+2 \right)\]

Note:
We should have a better knowledge in the topic of pre-calculus for solving this type of question easily. We should the following formulas:
Product rule of differentiation: \[d\left( u.v \right)=vd\left( u \right)-ud\left( v \right)\]
Logarithm formula: \[\ln {{x}^{a}}=a\ln x\]
\[d\left( \ln x \right)=\left( \dfrac{1}{x} \right)dx\]
\[d\left( {{x}^{n}} \right)=\left( n{{x}^{n-1}} \right)dx\]
\[\sqrt{x}={{x}^{\dfrac{1}{2}}}\]
\[\dfrac{1}{\sqrt{x}}={{x}^{-\left( \dfrac{1}{2} \right)}}\]
\[{{x}^{a}}{{x}^{b}}={{x}^{a+b}}\]