please help!!!!

Question

find y'' by implicit differentiation; √x + √y = 1

Answer 1

Hi Ashley,

Implicit differentiation is used here as a strategy without having to solve the equation directly for y first.

Starting with √x + √y = 1 --> Note that a square root is the same as raising it to an exponent of 1/2 (Numerator indicates the power, which is 1, and the denominator indicates the root, which is 2). If we rewrite it this way, it can help us differentiate using the power rule.

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

(1/2)x^-1/2 + (1/2)y^-1/2 (dy/dx) = 0 --> We are multiplying by dy/dx because this is the derivative of y (y') with respect to x. If we solve for dy/dx, we will get:

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

dy/dx = -y^1/2/x^1/2

Now, let's take the second derivative since we want to find y''. We can use the power rule again! Note that y' = dy/dx

y' = -y^1/2/x^1/2 --> We can use the quotient rule here, which states to the take (u'v - v'u)/v². Suppose -y^1/2 is u (numerator), and x^1/2 is v (denominator).

y'' = ((-1/2)y^-1/2 (dy/dx) * x^1/2 - (1/2)x^-1/2 * -y^1/2)

(x^1/2)²

> (x^1/2)² in the denominator will turn out to be just x.

> Additionally, we can multiply every term by 2 on the top and bottom so that we no longer have a fraction in the numerator (gets rid of the 1/2).

> We can substitute dy/dx with -y^1/2/x^1/2

With these three steps, this will now get us to:

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

(2x)

Note that the square root of y's and square root of x's cancel out in the first term and equals 1.

y'' = [1 + y^1/2/x^1/2]

2x

After using square root of x as the common denominator in the numerator, we get:

y'' = [(x^1/2 + y^1/2)/x^1/2]

2x

Remember that 1/2 exponent means square root. Also recall that √x + √y = 1 from the original equation. Therefore, this will simplify to:

y'' = [(√x + √y)/√x]

2x

y'' = 1/(2x*√x)

Let me know if you have any questions, and feel free to reach out if you would like to further discuss!

Thanks!

Answer 2

√x + √y = 1

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

(1/√y)(dy/dx) = -(1/√x)

So, dy/dx = -√y / √x

d²y / dx² = -{√x(1/(2√y))(dy/dx) - √y(1/(2√x)] / x

= [-(1/2)(√x/√y)(-√y/√x)] / x + [(1/2)(√y/√x)] / x = 1 / (2x) + √y / (2x√x) = (√x + √y) / (2x√x)

But, √x + √y = 1, so d²y/dx² = 1 / (2x√x)

Answer 3

sqrx +sqry = 1

sqry = 1-sqrx

square both sides

y = 1 -2sqrx + x

y' = -1/sqrx + 1 = -x^-.5 + 1

y" = (1/2)/x^(3/2) = .5x^-1.5

or = 1/2x^1.5

= 1/2xsqrx if you want to avoid the decimals

using any other method such as implicit differentiation should give you the same basic or equivalent answer in terms of x and y

sqrx + sqry = 1

.5/x^.5 + (.5/y^.5)y' = 0

y' = - .5x^-.5/.5y^-.5

y' = - sqr(y/x)

y" = -.5/sqr(y/x))(-y/x^2)y' = .5yy'/x^2(sqr(y/x))= -.5y/x^2

=-.5(1-2sqrx + x)/x^2 = -.5/x^2 +x^-1.5 -.5/x

= -1/2x^2 +1/xsqrx - 1/2x

go with the 1st answer, as the calculations are too tedious in the 2nd to avoid some mistake along the way