Question

Question #a4db9

Answer 1

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(y)` and use the following idea:

`dy/dx = 1/(dx/dy) = 1/(x')`, something that we can prove later.

from the formal definition

`dx/dy = lim_(h to 0) (f(y+h) - f(y))/(h)`

`= lim_(h to 0) (sin(y+h) - sin(y))/(h)`

`= lim_(h to 0) (siny cos h + cos y sin h - sin y)/(h)`

`= lim_(h to 0) (siny (cos h -1) + cos y sin h )/(h)`

`= sin y lim_(h to 0) color(blue)((cos h -1)/h) + cos y lim_(h to 0) color(red)((sin h )/(h)) qquad triangle`

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

`lim_(h to 0) color(blue)((cos h -1)/h) = 0` See here

`lim_(h to 0) color(red)((sin h )/(h)) = 1` See here

So `triangle` becomes:

`x' = dx/dy = cos y`

`implies y' = 1/ cos y = 1/ sqrt(1-x^2)`

Now to prove that `y' = 1/(x')` and vice versa

if we start with:

`y = f(x)`

`x = f^(-1) (y) = g(y)`

`implies y = f(g(y)) qquad triangle`

`d/dy(y) = 1 qquad square`

But by the chain rule applied to `triangle`

`dy/dy = d/dy( f(g(y))) = f'(x)g'(y) = y' x'`

So by `square`

`y' = 1/(x')` and `x' = 1/(y')`

That completes this approach, I think.

Answer 2

Refer to the Explanation Section below.

Explanation:

Let `y=f(x)=sin^-1x`

Now, `f'(x)=lim_(t to x) {f(t)-f(x)}/(t-x)`

`lim_(t to x) (sin^-1t-sin^-1x)/(t-x)`

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

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

`:. sin phi=t, and, sin theta=x." Also, as "t to x, phi to theta.`

Hence, in this new scenario,

`f'(x)=lim_(phi to theta) (phi-theta)/(sin phi-sin theta)`

`=lim_(phi to theta) (phi-theta)/{2cos((phi+theta)/2)sin((phi-theta)/2)}`

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

Here, we see that, the `1^(st)" limit is 1, because of "lim_(x to 0) sinx/x=1` and, the `2^(nd)" is "1/cos((theta+theta)/2)=1/costheta.`

Therefore, `f'(x)=1/costheta`

Now, `sin theta=x rArr cos theta=+-sqrt(1-x^2)`

But, `theta in [-pi/2,pi/2], cos theta > 0, i.e., cos theta=+sqrt(1-x^2)`.

Therefore, `dy/dx=f'(x)=1/sqrt(1-x^2)`.

Enjoy Maths!