Use the quotient rule for differentiation, (dy)/(dx) = (v du)/(dx) - u (dv)/(dx)v2, where u = e8x and v = 1+e4x.

You're on a roll — here are the solutions to your questions: 1. Solve (dy)/(dx) if y = e^8x1+e^4x when y=1 and x=0. Step 1: Use the quotient rule for differentiation, (dy)/(dx) = (v du)/(dx) - u (dv)/(dx)v^2, where u = e^8x and v = 1+e^4x. First, find the derivatives of u and v: (du)/(dx) = (d)/(dx)(e^8x) = 8e^8x (dv)/(dx) = (d)/(dx)(1+e^4x) = 4e^4x Step 2: Apply the quotient rule. (dy)/(dx) = (1+e^4x)(8e^8x) - (e^8x)(4e^4x)(1+e^4x)^2 Step 3: Simplify the expression. (dy)/(dx) = 8e^8x + 8e^12x - 4e^12x(1+e^4x)^2 (dy)/(dx) = 8e^8x + 4e^12x(1+e^4x)^2 Step 4: Evaluate (dy)/(dx) at x=0. Substitute x=0 into the derivative: (dy)/(dx)|_x=0 = 8e^8(0) + 4e^12(0)(1+e^4(0))^2 (dy)/(dx)|_x=0 = (8e^0 + 4e^0)/((1+e^0)^2) Since e^0 = 1: (dy)/(dx)|_x=0 = (8(1) + 4(1))/((1+1)^2) = (8+4)/(2^2) = (12)/(4) = 3 The condition y=1 at x=0 is not consistent with the given function y(0) = (e^0)/(1+e^0) = (1)/(2). We proceed by evaluating the derivative at x=0. The final answer is 3. 2. Solve (dy)/(dx) if y = ((x+1)/(x-1)). Step 1: Use the logarithm property ((A)/(B)) = A - B. y = (x+1) - (x-1) Step 2: Differentiate each term with respect to x. Recall that (d)/(dx)( u) = (1)/(u) (du)/(dx). (dy)/(dx) = (d)/(dx)((x+1)) - (d)/(dx)((x-1)) (dy)/(dx) = (1)/(x+1) · (d)/(dx)(x+1) - (1)/(x-1) · (d)/(dx)(x-1) (dy)/(dx) = (1)/(x+1)(1) - (1)/(x-1)(1) (dy)/(dx) = (1)/(x+1) - (1)/(x-1) Step 3: Combine the fractions by finding a common denominator. (dy)/(dx) = ((x-1) - (x+1))/((x+1)(x-1)) (dy)/(dx) = (x-1-x-1)/(x^2-1) (dy)/(dx) = (-2)/(x^2-1) The final answer is (-2)/(x^2-1). 3. Find intercepts and asymptotes for h(x) = (2x-4)/(x^3-1). a) x-intercepts: Set h(x) = 0. This occurs when the numerator is zero. 2x-4 = 0 2x = 4 x = 2 The x-intercept is (2, 0). b) y-intercepts: Set x = 0. h(0) = (2(0)-4)/(0^3-1) = (-4)/(-1) = 4 The y-intercept is (0, 4). c) Vertical asymptotes: Set the denominator to zero. x^3-1 = 0 Factor the difference of cubes: (x-1)(x^2+x+1) = 0. The real root is x-1=0 x=1. The quadratic factor x^2+x+1 has a discriminant = 1^2 - 4(1)(1) = -3 < 0, so it has no other real roots. Check that the numerator is not zero at x=1: 2(1)-4 = -2 0. The vertical asymptote is x=1. d) Horizontal asymptotes: Compare the degrees of the numerator and denominator. Degree of numerator (N) = 1. Degree of denominator (D) = 3. Since N < D, the horizontal asymptote is y=0. The horizontal asymptote is y=0. e) Slant asymptotes: A slant asymptote exists if the degree of the numerator is exactly one more than the degree of the denominator (N = D+1). In this case, N=1 and D=3, so N D+1. There is no slant asymptote. Got