Solve 80,82 and 83 and 85

here are the solutions for questions 80, 82, 83, and 85. 80. Find (dy)/(dx) if y = x^2 x. Step 1: Apply the product rule. The product rule states that if y = uv, then (dy)/(dx) = u'v + uv'. Let u = x^2 and v = x. Then u' = (d)/(dx)(x^2) = 2x and v' = (d)/(dx)( x) = x. (dy)/(dx) = (2x)( x) + (x^2)( x) Step 2: Simplify the expression. (dy)/(dx) = 2x x + x^2 x (dy)/(dx) = 2x x + x^2 x 82. Find (dy)/(dx) if y = (x^2)/(x+1). Step 1: Apply the quotient rule. The quotient rule states that if y = (f(x))/(g(x)), then (dy)/(dx) = (f'(x)g(x) - f(x)g'(x))/([g(x)]^2). Let f(x) = x^2 and g(x) = x+1. Then f'(x) = (d)/(dx)(x^2) = 2x and g'(x) = (d)/(dx)(x+1) = 1. (dy)/(dx) = ((2x)(x+1) - (x^2)(1))/((x+1)^2) Step 2: Simplify the numerator. (dy)/(dx) = (2x^2 + 2x - x^2)/((x+1)^2) (dy)/(dx) = (x^2 + 2x)/((x+1)^2) (dy)/(dx) = (x^2 + 2x)/((x+1)^2) 83. Find (dy)/(dx) if y = sqrt(x^2+1). Step 1: Rewrite the function and apply the chain rule. Rewrite y = sqrt(x^2+1) as y = (x^2+1)^1/2. The chain rule states that if y = [f(x)]^n, then (dy)/(dx) = n[f(x)]^n-1f'(x). Here, f(x) = x^2+1 and n = (1)/(2). So, f'(x) = (d)/(dx)(x^2+1) = 2x. (dy)/(dx) = (1)/(2)(x^2+1)^(1)/(2)-1 · (2x) (dy)/(dx) = (1)/(2)(x^2+1)^-(1)/(2) · (2x) Step 2: Simplify the expression. (dy)/(dx) = x(x^2+1)^-(1)/(2) (dy)/(dx) = (x)/(sqrt(x^2+1)) (dy)/(dx) = (x)/(sqrt(x^2+1)) 85. Find (dy)/(dx) if x^2 + y^2 = 25. Step 1: Differentiate both sides of the equation with respect to x using implicit differentiation. Remember that (d)/(dx)(y^2) = 2y (dy)/(dx) by the chain rule. (d)/(dx)(x^2) + (d)/(dx)(y^2) = (d)/(dx)(25) 2x + 2y (dy)/(dx) = 0 Step 2: Isolate (dy)/(dx). Subtract 2x from both sides: 2y (dy)/(dx) = -2x Divide by 2y: (dy)/(dx) = (-2x)/(2y) Step 3: Simplify the expression. (dy)/(dx) = -(x)/(y) (dy)/(dx) = -(x)/(y) What's next?