Question

If 3x2 + y2 = 7 then evaluate the second derivative of y with respect to x when x = 1 and y = 2. Round your answer to 2 decimal places. Use the hyphen symbol, -, for negative values.

Answer 1

2.63 to 2 DP's.

Explanation:

3x^2 + y^2 = 7

Finding the first derivative

6x + 2y . dy/dx = 0

3x + y.dy/dx = 0

dy/dx = -3x/y

Now we find the second derivative ( we apply the product rule):

3x + y.dy/dx = 0

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

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

Now we substitute for dy/dx in the above equation:

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) = -3 - 9x^2/y^2

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

d^2y/dx^2 = -3/y - 9x^2/ y^3

When x = 1 and y = 2:

d^2y/dx^2 = -3/2 - 9(1)^2 / 2^3

= -3/2 -9/8

= -2.625