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Differentiation of arctan(x), also written tan^(-1)(x), requires a little knowledge of regular trigonometric algebra and differentiation, and implicit differentiation.
Let's set y = arctan(x) to start. Then, by the definition the arctan function, tan(y) = x. We recall the derivative of the tan function, d(tan(u))/du = (sec(u))^2.
So, if we differentiate both sides of our equation with respect to x, we find that (dy/dx)((sec(y))^2) = 1, using the rules of implicit differentiation. Now, we can rearrange our equation for dy/dx = 1/((sec(y))^2).
Thinking about some trigonometric algebra, we know that* (sec(y))^2 = 1 + (tan(y))^2. Additionally, remember that tan(y) = x as we said earlier! Using this information, simple substitution back into our equation yields dy/dx = 1/(1+x^2).
So, d(arctan(x))/dx = 1/(1+x^2).
*For a reminder of where this comes from, think about the trigonometric relation: (sin(y))^2 + (cos(y))^2 = 1.see more