0
$\begingroup$

Given:

$[\tan^{-1}(x)]^2+[\cot^{-1}(y)]^2=1$

Find the tangent line equation to the graph at the point $(1,0)$ by implicit differentiation

I found the derivative:

$\dfrac{dy}{dx}=\dfrac{4\tan^{-1}(x)\cdot \cot^{-1}(y)}{(y^2+1)(x^2+1)}$

I may have done my derivative wrong, but my main concern is at some point $0$ will be plugged into $\cot$ inverse, resulting in division by zero.

I need help with this scenario.

$\endgroup$
3
  • $\begingroup$ Why do you say that evaluating the inverse cotangent function at 0 will result in a division by zero? $$\cot^{-1}(0) = \frac{\pi}{2}$$ $\endgroup$
    – SplitInfinity
    Commented Feb 27, 2016 at 4:39
  • $\begingroup$ You know I clearly mixed things up, thanks. $\endgroup$
    – User 210
    Commented Feb 27, 2016 at 4:52
  • $\begingroup$ $(\tan^{-1} 1)^{2}+(\cot^{-1} 0)^{2}=\frac{5\pi^{2}}{16} \neq 1$ $\endgroup$
    – Ng Chung Tak
    Commented Feb 27, 2016 at 9:25

1 Answer 1

Reset to default
0
$\begingroup$

$\displaystyle \frac{2\tan^{-1} x}{1+x^{2}} \, dx-\frac{2\cot^{-1} y}{1+y^{2}} \, dy=0$

$\displaystyle \frac{dy}{dx}=\frac{(1+y^{2})\tan^{-1} x}{(1+x^{2})\cot^{-1} y}$

Note that $(1,0)$ doesn't lie on the graph.

The graph can be parametrized as $(x,y)=(\tan \cos t,\cot \sin t)$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .