# How do you find #(d^2y)/(dx^2)# for #3x^2+y^2=2#?

##### 2 Answers

# (d^2y)/(dx^2) =- 6/y^3 #

#### Explanation:

When we differentiate

We have:

# 3x^2 + y^2 = 2 #

Differentiate wrt

# \ \ \ 6x + 2ydy/dx = 0 #

# :. 3x + ydy/dx = 0 #

And now (as we want the second derivative) we differentiate wrt x again, this time we must also apply the product rule:

# 3 + (y)((d^2y)/(dx^2)) + (dy/dx)(dy/dx) = 0 #

# :. 3 + y(d^2y)/(dx^2) + (dy/dx)^2 = 0 #

We can rearrange the 1st derivative equation:

# :. dy/dx = -(3x)/y #

And if we substitute this into the 2nd derivative equation we get:

# :. 3 + y(d^2y)/(dx^2) + (-(3x)/y)^2 = 0 #

# :. 3 + y(d^2y)/(dx^2) + (9x^2)/y^2 = 0 #

# :. y(d^2y)/(dx^2) =- (9x^2)/y^2 -3 #

# :. (d^2y)/(dx^2) =- (9x^2)/y^3 -3/y #

# " "=- (3(3x^2+y^2))/y^3 #

# " "=- (3(2))/y^3 #

# " "=- 6/y^3 #

#### Explanation:

We can solve the equation to make

So:

We can also proceed implicitly by differentiating both sides of the equation with respect to

Solve for

Differentiate again:

substituting

and as

simplifying:

and substituting