Question

What is the derivative of \[y={{\left( \sqrt{x} \right)}^{x}}\]?

Answer 1

Hint: In this question we have to use the concepts of derivatives of functions of a function along with the concept of logarithm. Here we have to use the rules of logarithm such as \[\ln \left( {{a}^{m}} \right)=m\ln \left( a \right)\]. Also we have to use the product rule of derivatives that is \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].

Complete step by step solution:
Now, we have to find the derivative of \[y={{\left( \sqrt{x} \right)}^{x}}\]
For this consider,
\[\Rightarrow y={{\left( \sqrt{x} \right)}^{x}}\]
Let us apply natural log on both sides
\[\Rightarrow \ln y=\ln \left[ {{\left( \sqrt{x} \right)}^{x}} \right]\]
As we know that, \[\ln \left( {{a}^{m}} \right)=m\ln \left( a \right)\] we get,
\[\Rightarrow \ln y=x\left[ \ln \left( \sqrt{x} \right) \right]\]
Now we take derivatives on both sides,
\[\Rightarrow \dfrac{d}{dx}\left( \ln y \right)=\dfrac{d}{dx}x\left[ \ln \left( \sqrt{x} \right) \right]\]
We know that \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\] and product rule of derivatives that is \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
\[\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=x\dfrac{d}{dx}\left[ \ln \left( \sqrt{x} \right) \right]+\ln \left( \sqrt{x} \right)\dfrac{d}{dx}\left( x \right)\]
\[\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=x\dfrac{1}{\sqrt{x}}\dfrac{d}{dx}\sqrt{x}+\ln \left( \sqrt{x} \right)\cdot 1\]
As we know the derivative of \[\sqrt{x}\] is equal to \[\dfrac{1}{2\sqrt{x}}\] that is \[\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}\]
\[\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{x}{\sqrt{x}}\cdot \dfrac{1}{2\sqrt{x}}+\ln \left( \sqrt{x} \right)\]
As \[\sqrt{x}\cdot \sqrt{x}={{\left( \sqrt{x} \right)}^{2}}={{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}=x\] we can write,
\[\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{x}{2x}+\ln \left( \sqrt{x} \right)\]
\[\Rightarrow \dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{2}+\ln \left( \sqrt{x} \right)\]
By taking y on right hand side we get,
\[\Rightarrow \dfrac{dy}{dx}=y\left[ \dfrac{1}{2}+\ln \left( \sqrt{x} \right) \right]\]
By substituting the value of y we can write,
\[\Rightarrow \dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left[ \dfrac{1}{2}+\ln \left( \sqrt{x} \right) \right]\]
As we know that, \[\sqrt{x}\] can be expressed as \[{{x}^{\dfrac{1}{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left[ \dfrac{1}{2}+\ln \left( {{x}^{\dfrac{1}{2}}} \right) \right]\]
By using \[\ln \left( {{a}^{m}} \right)=m\ln \left( a \right)\] we can write
\[\Rightarrow \dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left[ \dfrac{1}{2}+\dfrac{1}{2}\ln \left( x \right) \right]\]
\[\Rightarrow \dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left( \dfrac{1}{2} \right)\left( 1+\left( \ln x \right) \right)\]
Hence, the derivative of \[y={{\left( \sqrt{x} \right)}^{x}}\] is given by \[\dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left( \dfrac{1}{2} \right)\left( 1+\left( \ln x \right) \right)\]

Note: In this question one of the students may solve as follows:
\[\Rightarrow y={{\left( \sqrt{x} \right)}^{x}}\]
As we know that, \[\sqrt{x}\] can be expressed as \[{{x}^{\dfrac{1}{2}}}\]
\[\Rightarrow y={{\left( {{x}^{\dfrac{1}{2}}} \right)}^{x}}\]
Now by using \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\] we can write
\[\Rightarrow y={{x}^{\dfrac{x}{2}}}\]
Now applying natural log on both sides we get
\[\Rightarrow \ln y=\dfrac{x}{2}\ln x\]
And now by taking exponential of both sides we can write
\[\Rightarrow y={{e}^{\dfrac{x}{2}\ln x}}\]
Now we differentiate both sides
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\dfrac{x}{2}\ln x}}\dfrac{d}{dx}\left( \dfrac{x}{2}\ln x \right)\]
To find the derivative of the bracket we have to use product rule \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\dfrac{x}{2}\ln x}}\left( \dfrac{x}{2}\dfrac{d}{dx}\left( \ln x \right)+\left( \ln x \right)\dfrac{d}{dx}\left( \dfrac{x}{2} \right) \right)\]
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\dfrac{x}{2}\ln x}}\left( \dfrac{x}{2}\dfrac{1}{x}+\left( \ln x \right)\left( \dfrac{1}{2} \right) \right)\]
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\dfrac{x}{2}\ln x}}\left( \dfrac{1}{2}+\left( \ln x \right)\left( \dfrac{1}{2} \right) \right)\]
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\dfrac{x}{2}\ln x}}\left( \dfrac{1}{2} \right)\left( 1+\left( \ln x \right) \right)\]
Now, as we have \[y={{e}^{\dfrac{x}{2}\ln x}}\] and \[y={{\left( \sqrt{x} \right)}^{x}}\], we can replace \[{{e}^{\dfrac{x}{2}\ln x}}\] by \[{{\left( \sqrt{x} \right)}^{x}}\] and hence we can write
\[\Rightarrow \dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left( \dfrac{1}{2} \right)\left( 1+\left( \ln x \right) \right)\]
Hence the derivative of \[y={{\left( \sqrt{x} \right)}^{x}}\] is given by \[\dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left( \dfrac{1}{2} \right)\left( 1+\left( \ln x \right) \right)\]