Plz show step to step.

Question

If f(x) is a twice-differentiable function such that f(2) = 2 and dy/dx=6 times the square root of the quantity x squared + 3 times y squared end quantity, then what is the value of d squared y over dx squared at x = 2?

Answer 1

I rewrote your question as "If f(x) is a twice-differentiable function such that f(2) = 2 and dy/dx=6√(x²+3y²), then what is the value of d²y/dx² squared at x = 2?"

Given that dy/dx = 6√(x²+3y²), we can determine d²y/dx² by implicit differentiation:

d²y/dx² = (x²+3y²)'[6/(2√(x²+3y²)]

d²y/dx² = [2x+6y(dy/dx)][3/√(x²+3y²)]

d²y/dx² = [2x+6y(6√(x²+3y²))][3/√(x²+3y²)] <-- Substitute dy/dx = 6√(x²+3y²)

d²y/dx² = [2x+36y√(x²+3y²)][3/√(x²+3y²)]

Now, we can find the value of d²y/dx² at (2,2):

d²y/dx² = [2x+36y√(x²+3y²)][3/√(x²+3y²)]

d²y/dx² = [2(2)+36(2)√(2²+3(2)²)][3/√(2²+3(2)²)]

d²y/dx² = [4+72√(4+3(4))][3/√(4+3(4))]

d²y/dx² = [4+72√(4+12)][3/√(4+12)]

d²y/dx² = [4+72√16][3/√16]

d²y/dx² = [4+72(4)](3/4)

d²y/dx² = (4+288)(3/4)

d²y/dx² = (292)(3/4)

d²y/dx² = 219

Hope this helped!