Can anyone help me with this calculus question?

Question

How would you find y'' by implicit differentiation for this function:
sqrt x+ sqrt y=1? That is, the sqaure root of x plus the square root of y equals 1.

Answer 1

I can only do direct differentiation:
sqrt x+ sqrt y=1
y=(1-sqrt(x))^2
y'= -(1-sqrt(x))/sqrt(x)
= -(sqrt(x)-1)
y'' = -1/(2sqrt(x))

Answer 2

For simplicity you can:

sqrt x+ sqrt y=1

sqrt(y) = 1 - sqrt(x)

y = 1 - 2sqrt(x) + x

y' = - 1/sqrt(x) + 1

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

Or something along those lines...

Answer 3

the problem would be easier if you choose x=sin power 4 a and y=cos power 4 a ;
y double derivative w.r.t x can be defined as the ratio of y double derivative w.r.t a and x double derivative w.r.t a ;
result is (y-3(xy)^.5)/(x-3(xy)^0.5)

Answer 4

are you sure that that is a square root of y? you see i'm answering it but since y is in a square root i couldn't find the y'

Answer 5

d/dx(sqrt x) = 1/(2sqrt x)
d/dx(sqrt y) = y'(y prime) / (2sqrt y)

so then it would just be
1/(2sqrt x) + y'(y prime) / (2sqrt y) = 0
y'(y prime) / (2sqrt y) = -1/(2sqrt x)
y' = -2sqrt y / 2sqrt x
y' = -sqrt y / sqrt x

Answer 6

.5x^-.5 + .5x^-.5(y')=0 first derivative. solve for y' .
y'=-x^-.5y^.5.. Differentiate again to get y''

y''=.5x^1.5 - .5x^-1