Question

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

Answer 1

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 $\dfrac{dy}{dx}={{\left( \sqrt{x} \right)}^{x}}\left( \dfrac{1}{2} \right)\left( 1+\left( \ln x \right) \right)$