If $y = \arctan((2x-1)/(1+x-x^2))$, then $dy/dx$ at $x=1$ is equal to?

Question

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? You can check the question and the 2 approaches in the link below.

Question and 2 possible solutions

I have an exam tomorrow, so couldn't learn mathjax to type my question, would really help if you'd understand, thanks.

Answer 1

A computationally simpler method of directly calculating the derivative at $x = 1$ is to write $$\tan y = \frac{2x-1}{1+x-x^2}$$ so that $$\log \tan y = \log (2x-1) - \log (1 + x - x^2).$$ Then implicit differentiation yields $$\frac{\sec^2 y}{\tan y} \frac{dy}{dx} = \frac{2}{2x-1} - \frac{1 - 2x}{1 + x - x^2}.$$ Since, when $x = 1$, we have $\tan y = \frac{2 - 1}{1 + 1 - 1} = 1$, it follows from the trigonometric identity $\sec^2 y = 1 + \tan^2 y$ that $\sec^2 y = 1 + 1^2 = 2$; 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.

Answer 2

$$\frac{d}{dx}\tan^{-1}\frac{2x-1}{1+x-x^2}=\frac{1}{ {\left(1+\frac{(2x-1)^2}{(1+x-x^2)^2} \right)}}\frac{(1+x-x^2(2-(2x-1)(1-2x)}{(1+x-x^2)^2}$$ $$=\frac{2x^2-2x+3}{2-2x+3x^2-2x^3+x^4}$$

Answer 3

Your error is in the first approach. In the derivative of $$ y=\arctan(x)-\arctan(1-x)+k\pi $$ you missed the inner derivative of the second term, you should have gotten $$ \frac{dy}{dx}=\frac1{1+x^2}-\left(-\frac1{1+(1-x)^2}\right) $$ 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 (like there would be in computing the derivative directly).

Answer 4

Calculate the derivative directly using first principles.

$\frac{\rm{d}}{\rm{d}x}\left(\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)\right) = \lim\limits_{x \to 0}\ \frac{1}{x}\left(\tan^{-1}\left(\frac{2(x+1)-1}{1+(x+1)-(x+1)^2}\right)-\tan^{-1}(1)\right) = \lim\limits_{x \to 0}\ \frac{1}{x}\left(\tan^{-1}\left(\frac{2x+1}{1-x-x^2}\right)-\tan^{-1}(1)\right)$

Now we simplify the numerator using inverse $\tan(x)$ formula i.e. $\tan^{-1}(A)- \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB} \right)$ and get:

$\left(\tan^{-1}\left(\frac{2x+1}{1-x-x^2}\right)-\tan^{-1}(1)\right) = \tan^{-1}\left(\frac{x^2+3x}{-x^2+x+2} \right)$.

Therefore, $\frac{\rm{d}}{\rm{d}x}\left(\tan^{-1}\left(\frac{2x-1}{1+x-x^2}\right)\right)$at x= 1 is $\lim\limits_{x \to 0}\left(\frac{1}{x}\tan^{-1}\left(\frac{x^2+3x}{-x^2+x+2} \right)\right)$.

Solving the limit using L'Hopital rule, we get:

$\lim\limits_{x \to 0}\left(\frac{1}{x}\tan^{-1}\left(\frac{x^2+3x}{-x^2+x+2} \right)\right) =\lim\limits_{x \to 0} \frac{1}{1+\left(\frac{x^2+3x}{-x^2+x+2}\right)^2} = 1.$

So $1$ is the correct answer. Though the approach is a lengthy one, but works almost always.