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

If $y=\mathrm{arctan}\left(\frac{2x-1}{1+x-{x}^{2}}\right)$, then $\frac{dy}{dx}$ 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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bekiffen32
In the derivative of
$y=\mathrm{arctan}\left(x\right)-\mathrm{arctan}\left(1-x\right)+k\pi$
$\frac{dy}{dx}=\frac{1}{1+{x}^{2}}-\left(-\frac{1}{1+{\left(1-x\right)}^{2}}\right)$
which is the same as in the second approach.
Also note that the inverse tangent is an odd function, so that $\mathrm{arctan}\left(1-x\right)=-\mathrm{arctan}\left(x-1\right)$, so that there is no structural difference between the two approaches
###### Not exactly what you’re looking for?
Fallbasisz8
A computationally simpler method of directly calculating the derivative at x=1 is to write
$\mathrm{tan}y=\frac{2x-1}{1+x-{x}^{2}}$
so that
$\mathrm{log}\mathrm{tan}y=\mathrm{log}\left(2x-1\right)-\mathrm{log}\left(1+x-{x}^{2}\right)$
Then implicit differentiation yields
$\frac{{\mathrm{sec}}^{2}y}{\mathrm{tan}y}\frac{dy}{dx}=\frac{2}{2x-1}-\frac{1-2x}{1+x-{x}^{2}}$
Since, when x=1, we have $\mathrm{tan}y=\frac{2-1}{1+1-1}=1$, it follows from the trigonometric identity ; therefore
$2{\left[\frac{dy}{dx}\right]}_{x=1}=\frac{2}{2-1}-\frac{1-2}{1+1-1}=3$
hence the answer is 3/2. Note this solution uses two tactics; logarithmic differentiation and implicit differentiation, to make the calculation extremely simple.