Differentiate tan^-1 ((2x)/(1-x^2)) with respect to sin^-1 ((2x)/(1+x^2))?

Apr 11, 2017

$- 1$

Explanation:

Calling

$\left\{\begin{matrix}{y}_{1} = \arctan \left({f}_{1} \left(x\right)\right) \\ {y}_{2} = \arcsin \left({f}_{2} \left(x\right)\right)\end{matrix}\right.$

with

$\left\{\begin{matrix}{f}_{1} \left(x\right) = \frac{2 x}{1 - {x}^{2}} \\ {f}_{2} \left(x\right) = \frac{2 x}{1 + {x}^{2}}\end{matrix}\right.$

we need $\frac{{\mathrm{dy}}_{1}}{{\mathrm{dy}}_{2}}$

but

$\frac{{\mathrm{dy}}_{1}}{{\mathrm{dy}}_{2}} = \left(\frac{{\mathrm{dy}}_{1}}{\mathrm{dx}}\right) / \left(\frac{{\mathrm{dy}}_{2}}{\mathrm{dx}}\right) = \frac{{\mathrm{dy}}_{1}}{\mathrm{dx}} \frac{\mathrm{dx}}{{\mathrm{dy}}_{2}} = \frac{{\mathrm{dy}}_{1}}{{\mathrm{dy}}_{2}}$

and

$\left\{\begin{matrix}\frac{d}{\mathrm{dx}} \arctan \left({f}_{1} \left(x\right)\right) = \frac{f {'}_{1} \left(x\right)}{1 + {f}_{1}^{2} \left(x\right)} \\ \frac{d}{\mathrm{dx}} \arcsin \left({f}_{2} \left(x\right)\right) = \frac{f {'}_{2} \left(x\right)}{\sqrt{1 - {f}_{2}^{2} \left(x\right)}}\end{matrix}\right.$

note that

$\frac{d}{\mathrm{dx}} \arctan \left(g \left(x\right)\right) = \frac{d}{\mathrm{dg}} \arctan \left(g\right) \frac{\mathrm{dg}}{\mathrm{dx}}$ and
$\frac{d}{\mathrm{dx}} \arcsin \left(g \left(x\right)\right) = \frac{d}{\mathrm{dg}} \arcsin \left(g\right) \frac{\mathrm{dg}}{\mathrm{dx}}$

and

$\left\{\begin{matrix}\frac{d}{\mathrm{dg}} \arctan \left(g\right) = \frac{1}{1 + {g}^{2}} \\ \frac{d}{\mathrm{dg}} \arcsin \left(g\right) = \frac{1}{\sqrt{1 - {g}^{2}}}\end{matrix}\right.$

so after substituting

$\left\{\begin{matrix}f {'}_{1} \left(x\right) = \frac{2 \left(1 + {x}^{2}\right)}{{x}^{2} - 1} ^ 2 \\ f {'}_{2} \left(x\right) = \frac{2 \left(1 - {x}^{2}\right)}{1 + {x}^{2}} ^ 2\end{matrix}\right.$

and simplifying

$\frac{{\mathrm{dy}}_{1}}{{\mathrm{dy}}_{2}} = \frac{\frac{f {'}_{1} \left(x\right)}{1 + {f}_{1}^{2} \left(x\right)}}{\frac{f {'}_{2} \left(x\right)}{\sqrt{1 - {f}_{2}^{2} \left(x\right)}}} = - 1$

Apr 12, 2017

"The Reqd. Deri.="1, if -1ltxlt1;
$= - 1 , \mathmr{if} x > 1 :$
$- 1 , \mathmr{if} x < - 1.$

Explanation:

Let $u = {\tan}^{-} 1 \left(\frac{2 x}{1 - {x}^{2}}\right) , \mathmr{and} , v = {\sin}^{-} 1 \left(\frac{2 x}{1 + {x}^{2}}\right) .$

Note that, because of the Dr. of $u , x \in \mathbb{R} - \left\{\pm 1\right\} \ldots . \left(\ast\right)$

$\therefore x < - 1 , \mathmr{and} , - 1 < x < 1 , \mathmr{and} , x > 1.$

Subst. x=tantheta. because (ast), theta in (-pi/2,pi/2)-{+-pi/4}, &, theta=tan^-1x.

$\therefore u = {\tan}^{-} 1 \left\{\frac{2 \tan \theta}{1 - {\tan}^{2} \theta}\right\} = {\tan}^{-} 1 \left(\tan 2 \theta\right) , \mathmr{and} ,$

$v = {\sin}^{-} 1 \left\{\frac{2 \tan \theta}{1 + {\tan}^{2} \theta}\right\} = {\sin}^{-} 1 \left(\sin 2 \theta\right) .$

Case (1) : $- 1 < x < 0 , \mathmr{and} , 0 < x < 1.$

:. tan(-pi/4) lt tantheta lt tan 0, &, tan0 lt tantheta lt tan(pi/4).

Since, $\tan$ fun. is $\uparrow$ in all quadrants, it follows that,

$- \frac{\pi}{4} < \theta < 0 , \mathmr{and} , 0 < \theta < \frac{\pi}{4.}$

:. -pi/2 lt 2theta lt 0, &, 0 lt 2theta lt pi/2.

&, $\therefore$ by the Defns of ${\tan}^{-} 1 \mathmr{and} {\sin}^{-} 1$ functions, we have,

$u = {\tan}^{-} 1 \left(\tan 2 \theta\right) = 2 \theta , \mathmr{and} , v = {\sin}^{-} 1 \left(\sin 2 \theta\right) = 2 \theta .$

Thus, $u = 2 {\tan}^{-} 1 x = v , \mathmr{if} , - 1 < x < 1.$

$\text{Therefore, the Reqd. Deri.=} \frac{\mathrm{du}}{\mathrm{dv}} = \frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}} ,$

$= \frac{\frac{2}{1 + {x}^{2}}}{\frac{2}{1 + {x}^{2}}} = 1 , \mathmr{if} - 1 < x < 1.$

Case (2) : $x > 1.$

$\therefore \tan \theta > \tan \left(\frac{\pi}{4}\right) \Rightarrow \theta > \frac{\pi}{4.} . . \left[\because , \tan \text{ is "uarr}\right]$

Preferably, $\frac{\pi}{4} < \theta < \frac{\pi}{2} \Rightarrow \frac{\pi}{2} < 2 \theta < \pi .$

$\Rightarrow \frac{\pi}{2} - \pi < 2 \theta - \pi < \pi - \pi , i . e . , - \frac{\pi}{2} < 2 \theta - \pi < 0.$

Then, $\tan \left(2 \theta - \pi\right) = - \tan \left(\pi - 2 \theta\right) = - \left(- \tan 2 \theta\right) = \tan 2 \theta .$

$\therefore u = {\tan}^{-} 1 \left(\tan 2 \theta\right) = {\tan}^{-} 1 \left(\tan \left(2 \theta - \pi\right)\right) , \text{ where } \left(2 \theta - \pi\right) \in \left(- \frac{\pi}{2} , 0\right) \subset \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$

$\therefore$ by the Defns. of ${\tan}^{-} 1 \mathmr{and} {\sin}^{-} 1$ functions, we get,

u=2theta-pi=2tan^-1x-pi;

$\text{Also, } \sin \left(2 \theta - \pi\right) = - \sin \left(\pi - 2 \theta\right) = - \sin 2 \theta$

$\therefore \sin 2 \theta = - \sin \left(2 \theta - \pi\right) .$

$\therefore v = {\sin}^{-} 1 \left(\sin 2 \theta\right) = {\sin}^{-} 1 \left(- \sin \left(2 \theta - \pi\right)\right) = - {\sin}^{-} 1 \left(\sin \left(2 \theta - \pi\right)\right) = - \left(2 \theta - \pi\right) = \pi - 2 \tan \theta = \pi - 2 {\tan}^{-} 1 x , \left(x > 1\right)$

$\therefore \text{ The Reqd. Deri.=} \frac{\frac{2}{1 + {x}^{2}} - 0}{0 - \frac{2}{1 + {x}^{2}}} = - 1 , \mathmr{if} x > 1.$

Case (3) : $x < - 1.$

In this case, $x < - 1 \Rightarrow \theta < - \frac{\pi}{4.}$

We take, $- \frac{\pi}{2} < \theta < - \frac{\pi}{4} \therefore - \pi < 2 \theta < - \frac{\pi}{2.}$

$\therefore 0 < \left(\pi + 2 \theta\right) < \frac{\pi}{2} \Rightarrow \left(\pi + 2 \theta\right) \in \left(0 , \frac{\pi}{2}\right) \subset \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) .$

Also, tan(pi+2theta)=tan2theta, &, sin(pi+2theta)=-sin2theta.

$\therefore u = {\tan}^{-} 1 \left(\tan 2 \theta\right) = {\tan}^{-} 1 \left(\tan \left(\pi + 2 \theta\right)\right) = \pi + 2 \theta = \pi + 2 {\tan}^{-} 1 x ,$

and, $v = {\sin}^{-} 1 \left(\sin 2 \theta\right) = {\sin}^{-} 1 \left(- \sin \left(\pi + 2 \theta\right)\right) = - {\sin}^{-} 1 \left(\sin \left(\pi + 2 \theta\right)\right) = - \pi - 2 \theta = - \pi - 2 {\tan}^{-} 1 x , \left(x < - 1.\right)$

$\therefore \frac{\mathrm{du}}{\mathrm{dv}} = - 1 , x < - 1.$
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