At which value of x is the slope of the curve y=(1+x^2)^-1 the steepest

Since the slope at a certain point is the derivative, and by memory (1+x^2)^-1 =arctan(x), we just have to find arctan(x)'s highest point. Graphing it out based off of tan(x), we know that when x approaches π/2 in tan(x) then tan(x) approaches infinity, arctan(x) approaches π/2 when x approaches infinity. As arctan(x) gets closer to π/2 similar to how x gets closer to π/2 in tan(x) as well as that it gets a tiny bit higher (due to that it starts low) as x approaches infinity, your answer is infinity

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step-by-step explanation:

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