implicit differentiation?

Question

If y^2 - 3x=7, then [d^2y]/[dx^2] =
a)-6/7y^3
b)-3/y^3
c)3
d)3/2y
e)-9/4y^3
please explain because i don't fully understand how to do this

Answer 1

You want to find the second derivative, so first, solve for the first derivative, differentiating with respect to y:
y^2 - 3x=7
2y*dy/dx -3 =0
3/2y =dy/dx

Now we can find d^2y/dx^2 using the first derivative and the quotient rule:
quotient rule= f'[x]g[x] - f[x]g'[x]/(g[x])^2, where
f[x]= 3 // f'[x]= 0 // g[x]= 2y // g'[x]= 2*dy/dx

Plugging these values in, we'll get
=3/2y
= -3(2*dy/dx) / (2y^2) (note: but we know dy/dx = 3/2y)
= -3(2* (3/2y)) / (2y)^2
= -9/4y^3

[Answer: E]

Answer 2

y = f(x)

y² = f(x)²

d(y²)/dx = 2 f(x) f'(x) = 2 y dy/dx (this is the chain rule).

Now using the equation that defines y = f(x) (implicitly :),

y² - 3x = 7
2y(dy/dx) - 3 = 0 (taking d/dx of both sides)

dy/dx = 3/(2y)
d²y/dx² = [-3/(2y²)](dy/dx) = [-3/(2y²)][3/(2y)] = -9/(4y³)

Answer 3

y² - 3x = 7
y²- 3x - 7 = 0

y² = 3x + 7

y = √(3x + 7)

dy/dx = (3/2) * (3x + 7)^-(1/2)

d²y/dx² = -(9/4) (3x + 7)^-(3/2) =

-9/4(y^3)

Answer e)