What is the second derivative of #x + siny = 1# in terms of #x#?

If we start by solving for #y#, we get:

#siny = 1 - x#

#y = arcsin(1 - x)#

Now use the chain rule to find the first derivative. Let #y = arcsin(u)# and #u = 1 - x#. The derivative of #arcsinu# is #1/sqrt(1 - u^2)# and the derivative of #1 - x# is #-1#. By the chain rule, we have:

#y' = -1 * 1/sqrt(1 - u^2)#

#y' = -1/sqrt(1 - (1 - x)^2)#

#y' = -1/sqrt(1 - (1 - 2x + x^2))#

#y' = -1/sqrt(2x - x^2)#

#y' = -(2x - x^2)^(-1/2)#

Now we can find #y''# rather easily. Let #y = u^(-1/2)# with #u =2x -x^2#. Then #y' = -1/2u^(-3/2)# and #u' = 2 - 2x#.

#y'' = 1/2(2x - x^2)^(-3/2)(2 - 2x)#

Hopefully this helps!

Here's an alternate answer for an expression in #y# and then #x# using implicit differentiation, without requiring the memorization of the derivative of #arcsin#. I get:

#y'' = sec^2ytany#

#y'' = (1-x)/(2x - x^2)^"3/2"#

Given #x + siny = 1#, we will have that #y = arcsin(1-x)#. We will use that later, but for now, the implicit differentiation of the function gives:

#d/(dx)[x + siny] = d/(dx)[1]#

#=> 1 + cosy (dy)/(dx) = 0#,

remembering that #d/(dx)[f(y)] = (df)/(dy)(dy)/(dx)#.

As a result,

#y' -= (dy)/(dx) = -1/(cosy) = -secy#

And so, the second derivative is:

#y'' -= (d^2y)/(dx^2)#

#= d/(dx)[-secy] = -secytany (dy)/(dx)#

#= -secytany (-secy)#

This gives an expression in #bby# as:

#=> color(green)(y'' = sec^2ytany)#

Since #y = arcsin(1-x)#, we have:

#y'' = sec^2(arcsin(1-x))tan(arcsin(1-x))#

Here, #arcsin(1-x)# expresses an angle for the ratio #(1-x)/1#. Taking the #sec# of that, i.e. doing #sec(arcsin(1-x))#, returns a side-length ratio of a right triangle with a vertical leg length of #1 - x# and hypotenuse length of #1#.

The right triangle formed from the #arcsin# relationship has the remaining horizontal leg length given by:

#(1-x)^2 + ?^2 = 1^2#

#=> |?| = sqrt(1 - (1 - x)^2) = sqrt(2x - x^2)#

This means that, since #sec = 1/cos#, we take the ratio of the horizontal leg over the hypotenuse and reciprocate to get:

#sec(arcsin(1-x)) = 1/(cos(arcsin(1-x))) = 1/(sqrt(2x - x^2))#

Similarly, we take the #tan# of the angle given by #arcsin(1-x)# to get:

#tan(arcsin(1-x)) = (sin(arcsin(1-x)))/(cos(arcsin(1-x)))#

#= (1-x)/sqrt(2x - x^2)#

This then gives a final expression in #bbx# of:

#color(blue)(y'') = 1/(2x - x^2) cdot (1-x)/sqrt(2x - x^2) = color(blue)((1-x)/(2x - x^2)^"3/2")#