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