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

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CBSE Class 12 Maths Chapter-2 Inverse Trigonometric Functions Formula  Trigonometric Functions  Integration of Trigonometric Functions  Inverse Trigonometric Functions  Important Properties of Inverse Trigonometric Functions  Graphical Representation of Inverse Trigonometric Functions  CBSE Class 11 Maths Chapter 3 - Trigonometric Functions Formulas  Algebraic Expressions  Rational Expression  Numerical Expression  