# How do you differentiate  arctan(8^x)?

Sep 28, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{8}^{x} \left(\ln 8\right)}{{8}^{2 x} + 1}$

#### Explanation:

This kind of problem is usually done with implicit differentiation. We begin by writing the function in terms of $x$ and $y$ as such:

$y = \arctan \left({8}^{x}\right)$

Next, we use the definition of $\arctan$ to rewrite the inverse in terms of the original $\tan$:

$\tan y = {8}^{x} \text{ } \left[A\right]$

We can now proceed with the implicit differentiation with respect to $x$:

$\frac{d}{\mathrm{dx}} \left(\tan y\right) = \frac{d}{\mathrm{dx}} \left({8}^{x}\right)$
${\sec}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {8}^{x} \left(\ln 8\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{8}^{x} \left(\ln 8\right)}{{\sec}^{2} y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {8}^{x} \left(\ln 8\right) \cdot \left({\cos}^{2} y\right) \text{ } \left[B\right]$

In line [B] we note that the definition of $\sec y$ is $\frac{1}{\cos y}$ and make the suitable substitution.

The issue with this is the derivative still contains a term of $y$, which we must remove. The clever trick here is to go back to line [A] in the solution and recognize that we can use the trigonometric definition of tangent to define a right triangle using this information:

Note how the picture illustrates the fact that $\tan y = {8}^{x} = \left(\text{opposite")/("adjacent}\right)$ from trigonometry.

Using trigonometry, we can now evaluate what the value of $\cos y$ should be, and thus also ${\cos}^{2} y$ which needs to be replaced in our form of the derivative. Note that $c$ (the hypotenuse) can be derived from the Pythagorean Formula:

${c}^{2} = {a}^{2} + {b}^{2}$
${c}^{2} = {\left({8}^{x}\right)}^{2} + \left({1}^{2}\right)$
${c}^{2} = {8}^{2 x} + 1$

Thus:

$\cos y = \left(\text{adjacent")/("hypotenuse}\right) = \frac{1}{c}$
${\cos}^{2} y = {\left(\frac{1}{c}\right)}^{2} = \frac{1}{c} ^ 2 = \frac{1}{{8}^{2 x} + 1}$

Finally:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {8}^{x} \left(\ln 8\right) \cdot \left({\cos}^{2} y\right) = {8}^{x} \left(\ln 8\right) \cdot \left(\frac{1}{{8}^{2 x} + 1}\right)$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{8}^{x} \left(\ln 8\right)}{{8}^{2 x} + 1}$