How do you differentiate $\arctan \left( {{8}^{x}} \right)$?

Hint: From the question we are to differentiate $\arctan \left( {{8}^{x}} \right)$. For the questions of these types we will use implicit differentiation. We use the inverse trigonometry and rewrite the question and proceed the further solution or further differentiation by writing the function in terms of x and y as such.

Complete step-by-step solution:
Firstly, we will write the question in the terms of x and y. so, the expression will be reduced as follows.
$\Rightarrow y=\arctan \left( {{8}^{x}} \right)$
Now, from the definition of $\arctan $ we will rewrite the inverse function in the terms of original \[\tan \]. So, the equation will become as follows.
Let it be equation \[\left( 1 \right)\]
\[\Rightarrow \tan y={{8}^{x}}...\left( 1 \right)\]
Now, we will proceed the implicit differentiation to the above equation with respect to x. In implicit differentiation, we differentiate each side of the equation with two variables by treating one of the variables as function of the other.
So, we can now proceed with the implicit differentiation with respective to x as follows.
$\Rightarrow \dfrac{d}{dx}\left( \tan y \right)=\dfrac{d}{dx}\left( {{8}^{x}} \right)$
Here we use the formulae $\Rightarrow \dfrac{dy}{dx}\left( \tan x \right)={{\sec }^{2}}x$ and $\Rightarrow \dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\ln x$.
$\Rightarrow {{\sec }^{2}}y\dfrac{dy}{dx}=\left( {{8}^{x}} \right)\ln x$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{8}^{x}}\ln x}{{{\sec }^{2}}y}$
Here we will use the definition of $\sec x$ which is $\sec x=\dfrac{1}{\cos x}$ and make suitable substitution. So, the equation will be reduced as follows.
$\Rightarrow \dfrac{dy}{dx}={{8}^{x}}\ln x{{\cos }^{2}}y$
Here we will remove the y term by using the basic formula of trigonometry which is as follows.
From the above we know that,
$\Rightarrow \tan y={{8}^{x}}=\dfrac{opposite}{adjacent}$
From this we can write that,
$\Rightarrow \cos y=\dfrac{adjacent}{hypotenuse}=\dfrac{1}{\sqrt{{{8}^{2x}}+1}}$
Now, we will substitute the above got value in the equation. So, the equation will be simplified as follows.
$\Rightarrow \dfrac{dy}{dx}={{8}^{x}}\ln x{{\cos }^{2}}y$
$\Rightarrow \dfrac{dy}{dx}={{8}^{x}}\ln x{{\left( \dfrac{1}{\sqrt{{{8}^{2x}}+1}} \right)}^{2}}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{{{8}^{x}}\ln x}{{{8}^{2x}}+1}$
Therefore, the solution for the given question is $\Rightarrow \dfrac{dy}{dx}=\dfrac{{{8}^{x}}\ln x}{{{8}^{2x}}+1}$.

Note: Students must be very careful in doing the calculations. Students must have good knowledge in differentiation and trigonometry concepts. Students must not do mistake in using the formulae like,
$\Rightarrow \dfrac{dy}{dx}\left( \tan x \right)={{\sec }^{2}}x$
$\Rightarrow \dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\ln x$ and also the trigonometric formula like $\sec x=\dfrac{1}{\cos x}$ if we do any mistake it will make our whole solution wrong.