This kind of problem is usually done with implicit differentiation. We begin by writing the function in terms of xx and yy as such:
y = arctan(8^x)y=arctan(8x)
Next, we use the definition of arctanarctan to rewrite the inverse in terms of the original tantan:
tan y = 8^x" " [A]tany=8x [A]
We can now proceed with the implicit differentiation with respect to xx:
d/dx (tan y) = d/dx (8^x)ddx(tany)=ddx(8x)
sec^2 y * dy/dx = 8^x (ln 8)sec2y⋅dydx=8x(ln8)
dy/dx = (8^x (ln 8))/(sec^2 y)dydx=8x(ln8)sec2y
dy/dx = 8^x (ln 8) * (cos^2 y)" " [B]dydx=8x(ln8)⋅(cos2y) [B]
In line [B] we note that the definition of sec ysecy is 1/(cos y)1cosy and make the suitable substitution.
The issue with this is the derivative still contains a term of yy, 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")tany=8x=oppositeadjacent from trigonometry.
Using trigonometry, we can now evaluate what the value of cos ycosy should be, and thus also cos^2 ycos2y which needs to be replaced in our form of the derivative. Note that cc (the hypotenuse) can be derived from the Pythagorean Formula:
c^2 = a^2 + b^2 c2=a2+b2
c^2 = (8^x)^2 + (1^2) c2=(8x)2+(12)
c^2 = 8^(2x) + 1 c2=82x+1
Thus:
cos y = ("adjacent")/("hypotenuse") = 1/c cosy=adjacenthypotenuse=1c
cos^2 y = (1 / c)^2 = 1 / c^2 = 1 / (8^(2x) + 1) cos2y=(1c)2=1c2=182x+1
Finally:
dy/dx = 8^x (ln 8) * (cos^2 y) = 8^x (ln 8) * (1 / (8^(2x) + 1))dydx=8x(ln8)⋅(cos2y)=8x(ln8)⋅(182x+1)
:. dy/dx = (8^x (ln 8)) / (8^(2x) + 1)
