If sqrt(x) + sqrt(y) = 7 and y(25) = 4, find y' by implicit differentiation.?

Question

** Correction from perviously asked Question **

This is one of the questions on my Math Homework and really.. I don't even know how to attack this problem. The y(25) = 4 is throwing me off.

Does any one know how to solve and would please explain this problem to me?


Also: Most probably know but if not, sqrt(x) + sqrt(y) means the Square Root of x plus the Square Root of y.

Answer 1

sqrt(x) + sqrt(y) = 7
d/dx(sqrt(x) + sqrt(y)) = d/dx(7)
d/dx(sqrt(x)) + d/dx(sqrt(y) = 0
1/(2sqrt(x)) + y'/(2sqrt(y)) = 0
1 + y' * sqrt(x) / sqrt(y) = 0
y' * sqrt(x) / sqrt(y) = -1
-> y' = -sqrt(y / x)

Now sqrt(x) + sqrt(y) = 7, so
sqrt(y) = 7 - sqrt(x), and
y = (7 - sqrt(x))^2
= 49 - 14sqrt(x) + x

Plugging back in:
y' = -sqrt(y / x) = -sqrt((49 - 14sqrt(x) + x) / x)


The other piece is redundant, since we already know that sqrt(25) + sqrt(4) = 7, unless the question is asking for the _value_ of v' at x=25, in which case, plugging back into the arrowed equation:
y'(25) = -sqrt(y(25) / 25)
= -sqrt(4 / 25)
= -2 / 5

Answer 2

Never mind y(25) for now, just do implicit diff...

1 / (2sqrtx) + y' / (2sqrty) = 0
→ y' = -2sqrty / (2sqrtx) = (sqrtx - 7) / sqrtx = 1 - 7/sqrtx.

I think they throw in y(25)=4 just to tell you that you must use the positive square roots.

Answer 3

0 = d/dx [sqrt(x) + sqrt(y)] = 1/2sqrt(x) + y'/2sqrt(y)

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

y` = -sqrt(y)/sqrt(x)

y(25) = 4 thus sqrt(y)=2
and because the sum of the square root is 2 sqrt(x) = 5

and y' = -2/5

Answer 4

is y(25)= 4 the same as 25y=4

sqrt.x + sqrt.y=7
x^(1/2)+ y6(1/2)=7
dy/dx= (1/2)x^(-1/2) + (1/2)y^(-1/2)dy/dx
(1/2)y^(-1/2)dy/dx = -1/2x^(-1/2)
dy/dx= -1/2x^(-1/2) div. by 1/2y^(-1/2)

y=4/25 = 0.16
so sub. 016 in everywhere for y to find x i think.