![Find dy/dx at (5,-2) if 3x2-5y3+xy-1=0. Also find possible values of x for AB=BA given A=[[x2,3],[1,3x]] and B=[[3,6],[2,x]].](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1779308791017-e757370f69aa5285.png&w=3840&q=75)
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — here are the solutions to the problems from the image. Problem 3: If 3x^2-5y^3+xy-1=0, find (dy)/(dx) at the point (5,-2). Step 1: Differentiate the given equation implicitly with respect to x. (d)/(dx)(3x^2) - (d)/(dx)(5y^3) + (d)/(dx)(xy) - (d)/(dx)(1) = (d)/(dx)(0) Applying the power rule, chain rule, and product rule: 6x - 15y^2 (dy)/(dx) + (1 · y + x · (dy)/(dx)) - 0 = 0 6x - 15y^2 (dy)/(dx) + y + x (dy)/(dx) = 0 Step 2: Group terms containing (dy)/(dx) and terms not containing (dy)/(dx). (x - 15y^2)(dy)/(dx) = -6x - y Step 3: Solve for (dy)/(dx). (dy)/(dx) = (-6x - y)/(x - 15y^2) Step 4: Substitute the given point (x,y) = (5,-2) into the expression for (dy)/(dx). (dy)/(dx) = (-6(5) - (-2))/(5 - 15(-2)^2) (dy)/(dx) = (-30 + 2)/(5 - 15(4)) (dy)/(dx) = (-28)/(5 - 60) (dy)/(dx) = (-28)/(-55) (dy)/(dx) = (28)/(55) The value of (dy)/(dx) at the point (5,-2) is: (28)/(55) Problem 4: Given that A = x^2 & 3 \\ 1 & 3x and B = 3 & 6 \\ 2 & x , find the possible values of x such that AB = BA. Step 1: Calculate the product AB. AB = x^2 & 3 \\ 1 & 3x 3 & 6 \\ 2 & x AB = (x^2)(3) + (3)(2) & (x^2)(6) + (3)(x) \\ (1)(3) + (3x)(2) & (1)(6) + (3x)(x) AB = 3x^2 + 6 & 6x^2 + 3x \\ 3 + 6x & 6 + 3x^2 Step 2: Calculate the product BA. BA = 3 & 6 \\ 2 & x x^2 & 3 \\ 1 & 3x BA = (3)(x^2) + (6)(1) & (3)(3) + (6)(3x) \\ (2)(x^2) + (x)(1) & (2)(3) + (x)(3x) BA = 3x^2 + 6 & 9 + 18x \\ 2x^2 + x & 6 + 3x^2 Step 3: Set AB = BA and equate the corresponding elements. 3x^2 + 6 & 6x^2 + 3x \\ 3 + 6x & 6 + 3x^2 = 3x^2 + 6 & 9 + 18x \\ 2x^2 + x & 6 + 3x^2 Equating the top-left elements: 3x^2+6 = 3x^2+6 (This is always true). Equating the bottom-right elements: 6+3x^2 = 6+3x^2 (This is always true). Equating the top-right elements: 6x^2 + 3x = 9 + 18x 6x^2 + 3x - 18x - 9 = 0 6x^2 - 15x - 9 = 0 Divide by 3: 2x^2 - 5x - 3 = 0 Equating the bottom-left elements: 3 + 6x = 2x^2 + x 0 = 2x^2 + x - 6x - 3 0 = 2x^2 - 5x - 3 Both equations yield the same quadratic equation. Step 4: Solve the quadratic equation 2x^2 - 5x - 3 = 0. We can factor the quadratic equation: 2x^2 - 6x + x - 3 = 0 2x(x - 3) + 1(x - 3) = 0 (2x + 1)(x - 3) = 0 Set each factor to zero to find the values of x: 2x + 1 = 0 2x = -1 x = -(1)/(2) x - 3 = 0 x = 3 The possible values of x are: x = -(1)/(2) or x = 3 What's next?