Hi. I need help with implicit differentiation?

Question

Iam battling to find the second derivative using implicit differentiation of the following question:

sqrt(x) + sqrt(y) = 1


I know its an easy question and I know Iam making a mistake somewhere along the line. Please could anyone show me how to do it step by step. Many thanks

Answer 1

Okay:

Differentiate both sides:

.5x^(-1/2) + .5y^(-1/2)*dy/dx = 0

Mult. through by 2:

x^(-1/2) + y^(-1/2)dy/dx= 0, or

1/sqrtx+ (1/sqrt y)dy/dx = 0

dy/dx = - sqrt y / sqrt x

Now, at the very beginning, if we solve for y, we get

y = (1-sqrt x)^2. Substitute that in for y:

dy/dx = -(1-sqrt x)^2/sqrt x

If you multiply the top out:

dy/dx = (-1+ 2sqrt x - x)/sqrt x

Or dy/dx = -x^(-1/2) + 2 - x^1/2

Differentiating again, we get

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

Answer 2

From x^(1/2)+y^(1/2)=1, differentiate to get
(1/2)x^(-1/2) + (1/2)y^(-1/2) dy/dx =0.
Solve for dy/dx:
dy/dx=-x^(-1/2)y^(1/2).
Now differentiate again:
y''=(1/2)x^(-3/2) y^(1/2) -(1/2) x^(-1/2) y^(-1/2) dy/dx
=(1/2)x^(-3/2) y^(1/2) +(1/2) x^(-1)
=(1/2) x^(-3/2) [y^(1/2)+x^(1/2)]
=(1/2) x^(-3/2).
Where we used the known value of dy/dx in the second equality.

Answer 3

x^(1/2) + y^(1/2) = 1
First derivative:
(1/2)x^(-1/2) + (1/2)y^(-1/2) y' = 0
Subtract (1/2)y^(-1/2) y' from each side
(1/2)x^(-1/2) = -(1/2)y^(-1/2) y'
Divide each side by -(1/2)y^(-1/2)
(1/2)x^(-1/2)/[-(1/2)y^(-1/2)] = y'
y' = -sqrt(y)/sqrt(x)
Second derivative:
Rewrite y' as
y' = -y^(1/2) x^(-1/2)
Now use the product rule.

Answer 4

Wow, I havent been in school since May, and I really tried to do your problem, but I forgot how to do it now. :( I'm sorry.