# Question a4db9

Sep 4, 2016

See below

#### Explanation:

This is the first thing that occurs to me so a better answer may come along but I would operate on the basis that $x = \sin y = f \left(y\right)$ and use the following idea:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = \frac{1}{x '}$, something that we can prove later.

from the formal definition

$\frac{\mathrm{dx}}{\mathrm{dy}} = {\lim}_{h \to 0} \frac{f \left(y + h\right) - f \left(y\right)}{h}$

$= {\lim}_{h \to 0} \frac{\sin \left(y + h\right) - \sin \left(y\right)}{h}$

$= {\lim}_{h \to 0} \frac{\sin y \cos h + \cos y \sin h - \sin y}{h}$

$= {\lim}_{h \to 0} \frac{\sin y \left(\cos h - 1\right) + \cos y \sin h}{h}$

$= \sin y {\lim}_{h \to 0} \textcolor{b l u e}{\frac{\cos h - 1}{h}} + \cos y {\lim}_{h \to 0} \textcolor{red}{\frac{\sin h}{h}} q \quad \triangle$

The red and blue terms are well known limits provable using squeeze theorem.

${\lim}_{h \to 0} \textcolor{b l u e}{\frac{\cos h - 1}{h}} = 0$ See here

${\lim}_{h \to 0} \textcolor{red}{\frac{\sin h}{h}} = 1$ See here

So $\triangle$ becomes:

$x ' = \frac{\mathrm{dx}}{\mathrm{dy}} = \cos y$

$\implies y ' = \frac{1}{\cos} y = \frac{1}{\sqrt{1 - {x}^{2}}}$

Now to prove that $y ' = \frac{1}{x '}$ and vice versa

$y = f \left(x\right)$

$x = {f}^{- 1} \left(y\right) = g \left(y\right)$

$\implies y = f \left(g \left(y\right)\right) q \quad \triangle$

$\frac{d}{\mathrm{dy}} \left(y\right) = 1 q \quad \square$

But by the chain rule applied to $\triangle$

$\frac{\mathrm{dy}}{\mathrm{dy}} = \frac{d}{\mathrm{dy}} \left(f \left(g \left(y\right)\right)\right) = f ' \left(x\right) g ' \left(y\right) = y ' x '$

So by $\square$

$y ' = \frac{1}{x '}$ and $x ' = \frac{1}{y '}$

That completes this approach, I think.

Feb 7, 2017

Refer to the Explanation Section below.

#### Explanation:

Let $y = f \left(x\right) = {\sin}^{-} 1 x$

Now, $f ' \left(x\right) = {\lim}_{t \to x} \frac{f \left(t\right) - f \left(x\right)}{t - x}$

${\lim}_{t \to x} \frac{{\sin}^{-} 1 t - {\sin}^{-} 1 x}{t - x}$

To evaluate this Limit, let, sin^-1t=phi, &, sin^-1x=theta,

where, |t|,|x|<1; phi, theta in (-pi/2,pi/2).

$\therefore \sin \phi = t , \mathmr{and} , \sin \theta = x . \text{ Also, as } t \to x , \phi \to \theta .$

Hence, in this new scenario,

$f ' \left(x\right) = {\lim}_{\phi \to \theta} \frac{\phi - \theta}{\sin \phi - \sin \theta}$

$= {\lim}_{\phi \to \theta} \frac{\phi - \theta}{2 \cos \left(\frac{\phi + \theta}{2}\right) \sin \left(\frac{\phi - \theta}{2}\right)}$

=lim_(phi to theta){(phi-theta)/2}/(sin(phi-theta)/2)*1/(cos((phi+theta)/2)#

Here, we see that, the ${1}^{s t} \text{ limit is 1, because of } {\lim}_{x \to 0} \sin \frac{x}{x} = 1$ and, the ${2}^{n d} \text{ is } \frac{1}{\cos} \left(\frac{\theta + \theta}{2}\right) = \frac{1}{\cos} \theta .$

Therefore, $f ' \left(x\right) = \frac{1}{\cos} \theta$

Now, $\sin \theta = x \Rightarrow \cos \theta = \pm \sqrt{1 - {x}^{2}}$

But, $\theta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] , \cos \theta > 0 , i . e . , \cos \theta = + \sqrt{1 - {x}^{2}}$.

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$.

Enjoy Maths!