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

2 Answers
Feb 19, 2017

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

Explanation:

When we differentiate #y#, we get #dy/dx# When we differentiate a non implicit function of #y# then using the chain rule we can differentiate wrt #x# and multiply by #dy/dx#. When this is done in situ it is known as implicit differentiation.

We have:

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

Differentiate wrt #x# and we get:

# \ \ \ 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 #

Feb 19, 2017

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

Explanation:

We can solve the equation to make #y(x) # explicit:

#y(x) = +- sqrt(2-3x^2)#

So:

#dy/dx = +-(-3x)/(sqrt(2-3x^2))#

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

#(d^2y)/dx^2 = +- (-3(2-3x^2)-9x^2 )/((2-3x^2)sqrt(2-3x^2))#

#(d^2y)/dx^2 = +- (-6+9x^2-9x^2 )/((2-3x^2)sqrt(2-3x^2))#

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

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

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

#6x +2ydy/dx = 0#

Solve for #dy/dx#:

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

Differentiate again:

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

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

substituting #dy/dx# from the equation above:

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

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

and as #y^2 = 2-3x^2#:

#(d^2y)/dx^2 = -(9x^2 +6 -9x^2)/y^3#

simplifying:

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

and substituting #y# from the original equation we get the same result.