If y=arctan⁡(2x−11+x−x2), then dydx at x=1 is equal to? I have tried 2 different approaches, both yielding different results and both results are present in the options. Am I doing something wrong or why is this happening?

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If y=arctan⁡(2x−11+x−x2), then dydx at x=1 is equal to?  I have tried 2 different approaches, both yielding different results and both results are present in the options. Am I doing something wrong or why is this happening?

bekiffen32 · Accepted Answer

In the derivative of  y=arctan⁡(x)−arctan⁡(1−x)+kπ  dydx=11+x2−(−11+(1−x)2)  which is the same as in the second approach.   Also note that the inverse tangent is an odd function, so that arctan⁡(1−x)=−arctan⁡(x−1), so that there is no structural difference between the two approaches

Fallbasisz8 · Accepted Answer

A computationally simpler method of directly calculating the derivative at x=1 is to write  tan⁡y=2x−11+x−x2   so that   log⁡tan⁡y=log⁡(2x−1)−log⁡(1+x−x2)   Then implicit differentiation yields   sec2ytan⁡ydydx=22x−1−1−2x1+x−x2   Since, when x=1, we have tan⁡y=2−11+1−1=1, it follows from the trigonometric identity sec2y=1+tan2y  that  sec2y=1+12=2; therefore  2[dydx]x=1=22−1−1−21+1−1=3  hence the answer is 3/2. Note this solution uses two tactics; logarithmic differentiation and implicit differentiation, to make the calculation extremely simple.

If y=\arctan(\frac{2x−1}{1+x−x^2}), then \frac{dy}{dx} at x=1 is equal to? I have

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