# Differentiate the following expression ${{\left( x \right)}^{\tan x}}+{{\left( \tan x \right)}^{x}}$ with respect to $x$.

Answer

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Hint: Assume $u={{x}^{\tan x}}\ and\ v={{\left( \tan x \right)}^{x}}$ and differentiate the equations with respect to $x$ by taking log on both sides.

Complete step-by-step answer:

It is given in the question to differentiate the expression,${{\left( x \right)}^{\tan x}}+{{\left( \tan x \right)}^{x}}$ with respect to $x$.

Let us consider the given expression as,

$y={{x}^{\tan x}}+{{\left( \tan x \right)}^{x}}$

Let us assume that $u={{x}^{\tan x}}\ and\ v={{\left( \tan x \right)}^{x}}$ respectively, we get,

$y=u+v$

Now, we have $u={{x}^{\tan x}}$.

So, taking log on both the sides, we get,

$\log u=\log {{x}^{\tan x}}$

As we know that $\log {{a}^{b}}=b\log a$, we can write as,

$\Rightarrow \log u=\tan x.\log x$

Now, differentiating above equation with respect to $x$, we get,

\[\dfrac{d}{dx}\left( \log u \right)=\dfrac{d}{dx}\left( \tan x.\log x \right)\]

We can use the chain rule for differentiating the RHS as below,

\[\dfrac{d}{dx}\left( \log u \right)=\tan x\dfrac{d}{dx}\left( \log x \right)+\log x\dfrac{d}{dx}\left( \tan x \right)\]

Since, we know that the derivative of $\log x$is $\dfrac{1}{x}$ and $\tan x$ is ${{\sec }^{2}}x$, we can write,

$\begin{align}

& \Rightarrow \dfrac{1}{u}.\dfrac{dy}{dx}=\tan x\left( \dfrac{1}{x} \right)+\log x\left( {{\sec }^{2}}x \right) \\

& \Rightarrow \dfrac{dy}{dx}=u\left( \dfrac{\tan x}{x}+{{\sec }^{2}}x\log x \right)..............\left( 1 \right) \\

\end{align}$

Now putting the value of $u$ in equation (1), we get,

$\dfrac{du}{dx}={{x}^{\tan x}}\left( \dfrac{\tan x}{x}+{{\sec }^{2}}x\log x \right)$

Similarly, we have,$v={{\left( \tan x \right)}^{x}}$.

Again, taking log on both the sides, we get,

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

As we know that $\log {{a}^{b}}=b\log a$, we can write as,

$\log v=x\log \left( \tan x \right)$

Now, we will differentiate the above equation with respect to $x$. We can use the chain rule on RHS and the standard derivatives and we get,

$\begin{align}

& \Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\log \left( \tan x \right).1+x\left( \dfrac{1}{\tan x}.{{\sec }^{2}}x \right) \\

& \Rightarrow \dfrac{dv}{dx}=v\left( \log \left( \tan x \right)+\dfrac{x{{\sec }^{2}}x}{\tan x} \right) \\

& \Rightarrow \dfrac{dv}{dx}={{\left( \tan x \right)}^{x}}\left( \log \left( \tan x \right)+\dfrac{x{{\sec }^{2}}x}{\tan x} \right)............\left( 2 \right) \\

\end{align}$

Therefore, we get,

$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}.............\left( 3 \right)$

Now putting the value of $\dfrac{du}{dx}\ and\ \dfrac{dv}{dx}$in equation (3) we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx} \\

& \dfrac{dy}{dx}={{x}^{\tan x}}\left( \dfrac{\tan x}{x}+{{\sec }^{2}}x\log x \right)+{{\left( \tan x \right)}^{x}}\left( \log \left( \tan x \right)+\dfrac{x{{\sec }^{2}}x}{\tan x} \right) \\

\end{align}$

Note: If you don’t assume $u={{x}^{\left( \tan x \right)}}\ and\ v={{\left( \tan x \right)}^{x}}$ then the solution will become more complex. And the chances of error while solving it will also increase. Always replace using a small variable in place of the complex part of any equation.

Complete step-by-step answer:

It is given in the question to differentiate the expression,${{\left( x \right)}^{\tan x}}+{{\left( \tan x \right)}^{x}}$ with respect to $x$.

Let us consider the given expression as,

$y={{x}^{\tan x}}+{{\left( \tan x \right)}^{x}}$

Let us assume that $u={{x}^{\tan x}}\ and\ v={{\left( \tan x \right)}^{x}}$ respectively, we get,

$y=u+v$

Now, we have $u={{x}^{\tan x}}$.

So, taking log on both the sides, we get,

$\log u=\log {{x}^{\tan x}}$

As we know that $\log {{a}^{b}}=b\log a$, we can write as,

$\Rightarrow \log u=\tan x.\log x$

Now, differentiating above equation with respect to $x$, we get,

\[\dfrac{d}{dx}\left( \log u \right)=\dfrac{d}{dx}\left( \tan x.\log x \right)\]

We can use the chain rule for differentiating the RHS as below,

\[\dfrac{d}{dx}\left( \log u \right)=\tan x\dfrac{d}{dx}\left( \log x \right)+\log x\dfrac{d}{dx}\left( \tan x \right)\]

Since, we know that the derivative of $\log x$is $\dfrac{1}{x}$ and $\tan x$ is ${{\sec }^{2}}x$, we can write,

$\begin{align}

& \Rightarrow \dfrac{1}{u}.\dfrac{dy}{dx}=\tan x\left( \dfrac{1}{x} \right)+\log x\left( {{\sec }^{2}}x \right) \\

& \Rightarrow \dfrac{dy}{dx}=u\left( \dfrac{\tan x}{x}+{{\sec }^{2}}x\log x \right)..............\left( 1 \right) \\

\end{align}$

Now putting the value of $u$ in equation (1), we get,

$\dfrac{du}{dx}={{x}^{\tan x}}\left( \dfrac{\tan x}{x}+{{\sec }^{2}}x\log x \right)$

Similarly, we have,$v={{\left( \tan x \right)}^{x}}$.

Again, taking log on both the sides, we get,

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

As we know that $\log {{a}^{b}}=b\log a$, we can write as,

$\log v=x\log \left( \tan x \right)$

Now, we will differentiate the above equation with respect to $x$. We can use the chain rule on RHS and the standard derivatives and we get,

$\begin{align}

& \Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\log \left( \tan x \right).1+x\left( \dfrac{1}{\tan x}.{{\sec }^{2}}x \right) \\

& \Rightarrow \dfrac{dv}{dx}=v\left( \log \left( \tan x \right)+\dfrac{x{{\sec }^{2}}x}{\tan x} \right) \\

& \Rightarrow \dfrac{dv}{dx}={{\left( \tan x \right)}^{x}}\left( \log \left( \tan x \right)+\dfrac{x{{\sec }^{2}}x}{\tan x} \right)............\left( 2 \right) \\

\end{align}$

Therefore, we get,

$\dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx}.............\left( 3 \right)$

Now putting the value of $\dfrac{du}{dx}\ and\ \dfrac{dv}{dx}$in equation (3) we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{du}{dx}+\dfrac{dv}{dx} \\

& \dfrac{dy}{dx}={{x}^{\tan x}}\left( \dfrac{\tan x}{x}+{{\sec }^{2}}x\log x \right)+{{\left( \tan x \right)}^{x}}\left( \log \left( \tan x \right)+\dfrac{x{{\sec }^{2}}x}{\tan x} \right) \\

\end{align}$

Note: If you don’t assume $u={{x}^{\left( \tan x \right)}}\ and\ v={{\left( \tan x \right)}^{x}}$ then the solution will become more complex. And the chances of error while solving it will also increase. Always replace using a small variable in place of the complex part of any equation.

Last updated date: 25th Sep 2023

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