So find the second derivative of that same question

To find the second derivative (d^2y)/(dx^2) of the function y = (2x^3+4)/((5x-6)^2), we will differentiate the first derivative, which we found to be (dy)/(dx) = (10x^3 - 36x^2 - 40)/((5x-6)^3). Step 1: Identify U and V for the second derivative and find their derivatives. Let U = 10x^3 - 36x^2 - 40. Let V = (5x-6)^3. Find the derivative of U: U' = (d)/(dx)(10x^3 - 36x^2 - 40) = 30x^2 - 72x Find the derivative of V using the chain rule: V' = (d)/(dx)((5x-6)^3) Let w = 5x-6, so V = w^3. Then (dV)/(dw) = 3w^2 and (dw)/(dx) = 5. V' = (dV)/(dw) · (dw)/(dx) = 3(5x-6)^2 · 5 = 15(5x-6)^2 Step 2: Apply the quotient rule to find (d^2y)/(dx^2). The quotient rule states that if y = (U)/(V), then (d^2y)/(dx^2) = (U'V - UV')/(V^2). Substitute U, V, U', V' into the formula: (d^2y)/(dx^2) = ((30x^2 - 72x)(5x-6)^3 - (10x^3 - 36x^2 - 40)(15(5x-6)^2))/(((5x-6)^3)^2) (d^2y)/(dx^2) = ((30x^2 - 72x)(5x-6)^3 - 15(10x^3 - 36x^2 - 40)(5x-6)^2)/((5x-6)^6) Step 3: Simplify the expression. Factor out (5x-6)^2 from the numerator: (d^2y)/(dx^2) = ((5x-6)^2[(30x^2 - 72x)(5x-6) - 15(10x^3 - 36x^2 - 40)])/((5x-6)^6) Cancel (5x-6)^2 from the numerator and denominator: (d^2y)/(dx^2) = ((30x^2 - 72x)(5x-6) - 15(10x^3 - 36x^2 - 40))/((5x-6)^4) Step 4: Expand and combine terms in the numerator. Expand the first product: (30x^2 - 72x)(5x-6) = 30x^2(5x) + 30x^2(-6) - 72x(5x) - 72x(-6) = 150x^3 - 180x^2 - 360x^2 + 432x = 150x^3 - 540x^2 + 432x Expand the second product: -15(10x^3 - 36x^2 - 40) = -150x^3 + 540x^2 + 600 Combine these two expanded parts for the numerator: (150x^3 - 540x^2 + 432x) + (-150x^3 + 540x^2 + 600) = (150x^3 - 150x^3) + (-540x^2 + 540x^2) + 432x + 600 = 432x + 600 Factor out the common factor from the numerator (24): 432x + 600 = 24(18x + 25) Step 5: Write the final expression for the second derivative. (d^2y)/(dx^2) = (24(18x + 25))/((5x-6)^4) 3 done, 2 left today. You're making progress.