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# How do you differentiate ${{e}^{\tan x}}$?

Last updated date: 14th Jun 2024
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Hint: Assume the exponent of e as $f\left( x \right)$ to get the form of expression as ${{e}^{f\left( x \right)}}$. Now, differentiate the function with respect to the variable x and use the chain rule of differentiation to find the derivative of ${{e}^{f\left( x \right)}}$. First differentiate ${{e}^{f\left( x \right)}}$ with respect to $f\left( x \right)$ and then differentiate $f\left( x \right)$ with respect to x, which is given mathematically as: - $\dfrac{d\left[ {{e}^{f\left( x \right)}} \right]}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$. Use the basic formulas: - $\dfrac{d\left[ {{e}^{x}} \right]}{dx}={{e}^{x}}$ and $\dfrac{d\left[ \tan x \right]}{dx}={{\sec }^{2}}x$, to get the answer.

Complete step-by-step solution:
Here, we have been provided with the exponential function ${{e}^{\tan x}}$ and we are asked to find its derivative. Let us assume this function as y, that means we have to find the value of $\dfrac{dy}{dx}$.
$\because y={{e}^{\tan x}}$
Here, we can see that the exponent of e is also a function, so let us assume it as $f\left( x \right)$. So, we have the function given as: -
$\Rightarrow y={{e}^{f\left( x \right)}}$
Differentiating both the sides with respect to the variable x, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{e}^{f\left( x \right)}} \right]}{dx}$
Using the chain rule of differentiation, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{e}^{f\left( x \right)}} \right]}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$
What we are doing is, we are first differentiating ${{e}^{f\left( x \right)}}$ with respect to the function $f\left( x \right)$ and then we are differentiating the function $f\left( x \right)$ with respect to the variable x and considering their product. So, substituting the assumed function $\tan x=f\left( x \right)$, we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{e}^{\tan x}} \right)}{d\left( \tan x \right)}\times \dfrac{d\left( \tan x \right)}{dx}$
Using the formulas: - $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}$ and $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$, we get,
\begin{align} & \Rightarrow \dfrac{dy}{dx}={{e}^{\tan x}}\times {{\sec }^{2}}x \\ & \Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}x.{{e}^{\tan x}} \\ \end{align}
Hence, the derivative of ${{e}^{\tan x}}$ is ${{\sec }^{2}}x.{{e}^{\tan x}}$.

Note: One may note that there can be another method also to solve the question. What we can do is, we will take natural log, i.e., log to the base e, both the sides and simplify the R.H.S. by using the logarithmic formula: - $\log {{a}^{m}}=m\log a$. Now, we will differentiate both the sides with respect to x and use the chain rule in the L.H.S. to differentiate $\ln y$ to get the answer. We will use the formula $\dfrac{d\left( \ln x \right)}{dx}=\dfrac{1}{x}$. You must remember all the basic rules of differentiation like: - the product rule, chain rule, $\dfrac{u}{v}$ rule etc.