Finding Y prime and Y double prime using Implicit Differentiation

[Kstahl]

The question is," Find Y double prime by using implicit differentiation: x^1/2+y^1/2= 1

I found that Y prime= (-1/2(x^1/2)/1/2(y)^1/2)) but that is very ugly and much too difficult to use in order to find y double prime. How can I simplify it better and how do I find Y double prime because I still cannot even finish it it's too complicated. Thank you so much and sorry if my computer notation is confusing!

-Kstahl

 

[stapel]

Sometimes life is messy. That's okay. You just have to plow on.

 

[Deleted member 4993]

The question is," Find Y double prime by using implicit differentiation: x^1/2+y^1/2= 1

I found that Y prime= (-1/2(x^1/2)/1/2(y)^1/2)) but that is very ugly and much too difficult to use in order to find y double prime. How can I simplify it better and how do I find Y double prime because I still cannot even finish it it's too complicated. Thank you so much and sorry if my computer notation is confusing!

-Kstahl

If you inserted that expression in computer - you'll get a syntax error. You have "unbalanced" parenthesis. In any language - human or computer - it is fraught with problems.

x^(1/2) + y^(1/2) = 1

(1/2)*x^(-1/2) + (1/2) * y^(-1/2) * y' = 0

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

y" = [(1/2) * y^(-1/2) * y' * x - (1/2) * x^(-1/2) * y]/x

y" = [(1/2) * y^(-1/2) * {- (y/x)^(1/2)}* x - (1/2) * x^(-1/2) * y]/x

y" = -(1/2) * x^(-1/2)] * { x + y}/x = -(y + x)/[2*x^(3/2)]

Not so messy after all.....

 

[Ishuda]

The question is," Find Y double prime by using implicit differentiation: x^1/2+y^1/2= 1

I found that Y prime= (-1/2(x^1/2)/1/2(y)^1/2)) but that is very ugly and much too difficult to use in order to find y double prime. How can I simplify it better and how do I find Y double prime because I still cannot even finish it it's too complicated. Thank you so much and sorry if my computer notation is confusing!

-Kstahl

Or, another way:
Simplifying what you have (meant), we have
y' = -$\displaystyle (\frac{y}{x})^{\frac{1}{2}}$
we have
y'' =-$\displaystyle \frac{1}{2}(\frac{y}{x})^{-\frac{1}{2}}(\frac{y}{x})'=\frac{1}{2y'}(\frac{y}{x})'$
$\displaystyle (\frac{y}{x})' = \frac{y'}{x} - \frac{y}{x^2} = \frac{y'}{x} - \frac{y}{x}\frac{1}{x}=\frac{y'}{x} - (y')^2\frac{1}{x}=\frac{y'}{x}(1-y')$
So
y'' = $\displaystyle \frac{1}{2x}(1-y')$