Question

How do you find the derivative of $y={{x}^{\tan x}}$ ?

Answer 1

We first take natural logarithms on both sides of ths given equation. Thereafter, we perform the differentiation on both sides of the equation with respect to $x$ . Applying chain rule to the left hand side and product rule to the right hand side, and finally substituting the value of $y$ , we get our final answer.

Complete step-by-step answer:
The given expression is
$y={{x}^{\tan x}}$
We see that on the right hand side of the equation, the expression is of the form ${{\left( \text{variable} \right)}^{\text{variable}}}$ . So, directly we cannot differentiate it. Thus, we take natural logarithms on both sides of the given equation. The equation thus becomes,
$\Rightarrow \ln y=\ln \left( {{x}^{\tan x}} \right)$
Applying the property of logarithms that $\ln \left( {{a}^{b}} \right)=b\ln a$ in the above equation, we get,
$\Rightarrow \ln y=\left( \tan x \right)\ln x$
We now differentiate both sides of the above equation with respect to $x$ . We apply the chain rule of differentiation to the left hand side of the equation. The chain rule of differentiation states that
$\dfrac{d}{dx}\left( g\left( f\left( x \right) \right) \right)=\dfrac{dg}{df}\times \dfrac{df}{dx}$
The equation thus becomes,
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \left( \tan x \right)\ln x \right)$
The right hand side of the above equation is differentiated according to the product rule of differentiation which states that
$\dfrac{d}{dx}\left( u\times v \right)=u\times \dfrac{dv}{dx}+v\times \dfrac{du}{dx}$
The equation thus becomes,
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\tan x\times \dfrac{d\left( \ln x \right)}{dx}+\ln x\times \dfrac{d\left( \tan x \right)}{dx}$
Carrying on the differentiation, we get,
$\Rightarrow \dfrac{1}{y}\times \dfrac{dy}{dx}=\tan x\times \dfrac{1}{x}+\ln x\times {{\sec }^{2}}x$
Multiplying $y$ on both sides of the above equation, the equation thus becomes,
$\Rightarrow \dfrac{dy}{dx}=\left( \tan x\times \dfrac{1}{x}+\ln x\times {{\sec }^{2}}x \right)\times y$
Putting the value of $y={{x}^{\tan x}}$ given in the question, the equation thus becomes
$\Rightarrow \dfrac{dy}{dx}=\left( \tan x\times \dfrac{1}{x}+\ln x\times {{\sec }^{2}}x \right)\times {{x}^{\tan x}}$
Further simplification of the above equation cannot be done.
Therefore, we can conclude that the derivative of $y={{x}^{\tan x}}$ is $\left( \tan x\times \dfrac{1}{x}+\ln x\times {{\sec }^{2}}x \right)\times {{x}^{\tan x}}$ .

Note: The given problem requires various rules of differentiation, so we must be very careful while applying them. Students often make mistakes in the chain rule as it requires continuous differentiation of the functions.