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$y=x\log x+{{\left[ \log x \right]}^{x}}$

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We are given a function of x and we need to find the differentiation of that.

The function is $x\log x+{{\left[ \log x \right]}^{x}}$

Let’s separate the left side of the addition as the first term and the right-hand side as the second term.

Let’s start with the first term and differentiate it.

We have two functions multiplied to each other.

We need to use the product rule of differentiation.

The product rule says that if we have two functions $u\left( x \right)$ and $v\left( x \right)$ then, the differentiation of $y=u\left( x \right)\times v\left( x \right)$ is given by $\dfrac{dy}{dx}=u\left( x \right)\dfrac{dv\left( x \right)}{dx}+v\left( x \right)\dfrac{du\left( x \right)}{dx}$ .

In this case, let $u\left( x \right)=x$ and $v\left( x \right)=\log x$ .

Substituting the taking the differentiation we get,

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( x\log x \right)$

The differentiation of x is one, and the differentiation of log x is $\dfrac{1}{x}$ .

Substituting the values, we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{d}{dx}\left( x\log x \right) \\

& =x\dfrac{d\left( \log x \right)}{dx}+\log x\dfrac{dx}{dx} \\

& =x\times \dfrac{1}{x}+\log x \\

& =1+\log x \\

\end{align}$

Therefore, the value of the first term is $1+\log x$ .

Now, let’s find the differentiation of the second term.

we have let $t={{\left[ \log x \right]}^{x}}$.

Taking log on both sides we get,

$\log t=\log \left( {{\left( \log x \right)}^{x}} \right)$

There is a property where $\log {{a}^{b}}=b\log a$ using this we get,

$\log t=x\log \left( \log x \right)$

Now let’s differentiation the left-hand side first

We need to use chain rule for this,

Chain rule says that, $\dfrac{dy}{dt}=\dfrac{d}{dt}\left( g\left( f\left( t \right) \right) \right)=\dfrac{d\left( g\left( f\left( t \right) \right) \right)}{df\left( t \right)}\times \dfrac{df\left( t \right)}{dt}$

Using this we get,

$\dfrac{d\left( \log t \right)}{dx}=\dfrac{d\log t}{dt}\times \dfrac{dt}{dx}$

Solving this we get,

$\begin{align}

& \dfrac{d\left( \log t \right)}{dx}=\dfrac{d\log t}{dt}\times \dfrac{dt}{dx} \\

& =\dfrac{1}{t}\dfrac{dt}{dx} \\

\end{align}$

Now let's do the right-hand side,

$\dfrac{d}{dx}\left( x\log \left( \log x \right) \right)$

We again need to use the product rule of multiplication.

Let $u\left( x \right)=x,v\left( x \right)=\log \left( \log x \right)$ , by substituting we get,

$\begin{align}

& \dfrac{d}{dx}\left( x\log \left( \log x \right) \right)=\log \left( \log x \right)\dfrac{dx}{dx}+x\dfrac{d}{dx}\log \left( \log x \right) \\

& =\log \left( \log x \right)+x\dfrac{d}{dx}\log \left( \log x \right)

\end{align}$

Again, we need to use the chain rule, we get,

$\begin{align}

& \dfrac{d}{dx}\left( x\log \left( \log x \right) \right)=\log \left( \log x \right)\dfrac{dx}{dx}+x\dfrac{d}{dx}\log \left( \log x \right) \\

& =\log \left( \log x \right)+x\dfrac{d}{dx}\log \left( \log x \right) \\

& =\log \left( \log x \right)+x\left[ \dfrac{1}{\log x}\times \dfrac{1}{x} \right]

\end{align}$

Simplifying this and equation with the left-hand side we get,

$\dfrac{1}{t}\dfrac{dt}{dx}=\log \left( \log x \right)+\dfrac{1}{\log x}$

But we know that $t={{\left[ \log x \right]}^{x}}$ ,

Hence, we get,

$\dfrac{dt}{dx}={{\left[ \log x \right]}^{x}}\left[ \log \left( \log x \right)+\dfrac{1}{\log x} \right]$

We also have the addition rule of differentiation we have,

If we have $y=a\left( x \right)+b\left( x \right)$ , then $\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}$ .

Therefore, we just have to combine the first and second term.

By doing this we get,

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