Question #a4db9

2 Answers
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(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.

Feb 7, 2017

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!