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

stropa0u

stropa0u

Answered question

2022-01-24

If y=arctan(2x11+xx2), 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?

Answer & Explanation

bekiffen32

bekiffen32

Beginner2022-01-25Added 8 answers

In the derivative of
y=arctan(x)arctan(1x)+kπ
dydx=11+x2(11+(1x)2)
which is the same as in the second approach.
Also note that the inverse tangent is an odd function, so that arctan(1x)=arctan(x1), so that there is no structural difference between the two approaches
Fallbasisz8

Fallbasisz8

Beginner2022-01-26Added 9 answers

A computationally simpler method of directly calculating the derivative at x=1 is to write
tany=2x11+xx2
so that
logtany=log(2x1)log(1+xx2)
Then implicit differentiation yields
sec2ytanydydx=22x112x1+xx2
Since, when x=1, we have tany=211+11=1, it follows from the trigonometric identity sec2y=1+tan2y  that  sec2y=1+12=2; therefore
2[dydx]x=1=221121+11=3
hence the answer is 3/2. Note this solution uses two tactics; logarithmic differentiation and implicit differentiation, to make the calculation extremely simple.

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