implicit differentiation: Find 2nd derivative of y^2=x^3?

Question

i got the first derivative is
y'=(3x^2)/(2y)
Then i think you do quotient rule on the right side, but what do you do on the left?

Answer 1

y^2 = 3 x^3 ==> 2 y y' = 3 x^2 ==> y' = 3 x^2 / (2y) OK

Second derivative:

(y')' = y'' (by definition)

(3x^2) / (2y))' = [(6x)(2y)-(3x^2)(2y')] / (2y)^2 =
= [(6x)(2y)-(3x^2)(2(3 x^2 / (2y)))] / (2y)^2 =
= [12xy - 9x^4 / y] / (4 y^2)

Answer 2

y^2=x^3

2y(dy/dx) = 3x^2

y' = (3x^2) / 2y ........ (O.K. you got this far)

you are right to use the quotient rule on the right side.
For the left side you take the derivative of y', which is y''.

then you get what you are looking for, y''(x).

y"(x) = [12xy - 6y'x^2] / (4y^2)

now, you substitute y and y' into this expression for y"(x).

Answer 3

2yy'=3x^2 You got this right.
y'(2y')+(2y)y''=6x I suppose you could use quotient rule
So 2(y')^2 + 2yy'' =6x but I wanted to differentiate
So y'' = 3x/((y')^2+y) implicitly a second time

I hope that this helps.

Answer 4

The left side just becomes y'' = the second derivative

Answer 5

(x^2)y+(y^2)x=-2 differentiate the two components with appreciate to x 2x y + x^2 dy/dx + (2y dy/dx) x + y^2 (a million) = 0 2xy +x^2 dy/dx + 2xy dy/dx +y^2 = 0 dy/dx( x^2+2xy) = -(y^2+2xy) dy/dx = -(y^2+2xy) / (x^2+2xy) to discover the by-product at (-a million,2): substitute x=-a million and y=2 into dy/dx dy/dx = -(4-4) / (a million-4) = 0/-3 = 0

Answer 6

Then you can pretend 3x^2/2y = f[x]
Find the derivative of that...
=(6x)(2y) - (3x^2)(2)*dy/dx
_____________________
(2y)^2
But you know dy/dx = 3x^2/2y, so plug dy/dx into that and that'll be your d2y/dx^2

Answer 7

that's wrong:

you rearrange it to:

y=SQRT^2(x^3)

so

y= x^(3/2)

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

y'' = (3/4)*x^(-1/2)