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(8^x)
Next, we use the definition of arctan to rewrite the inverse in terms of the original tan:
tan y = 8^x" " [A]
We can now proceed with the implicit differentiation with respect to x:
d/dx (tan y) = d/dx (8^x)
sec^2 y * dy/dx = 8^x (ln 8)
dy/dx = (8^x (ln 8))/(sec^2 y)
dy/dx = 8^x (ln 8) * (cos^2 y)" " [B]
In line [B] we note that the definition of sec y is 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:
![enter image source here]()
Note how the picture illustrates the fact that tan y = 8^x = ("opposite")/("adjacent") 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 = (8^x)^2 + (1^2)
c^2 = 8^(2x) + 1
Thus:
cos y = ("adjacent")/("hypotenuse") = 1/c
cos^2 y = (1 / c)^2 = 1 / c^2 = 1 / (8^(2x) + 1)
Finally:
dy/dx = 8^x (ln 8) * (cos^2 y) = 8^x (ln 8) * (1 / (8^(2x) + 1))
:. dy/dx = (8^x (ln 8)) / (8^(2x) + 1)
