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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Here are the derivatives for the remaining functions: 6. (x + y) = y^2 x Step 1: Differentiate both sides with respect to x. Remember to use the chain rule for terms involving y and the product rule for y^2 x. (d)/(dx)((x + y)) = (d)/(dx)(y^2 x) (x + y) · (d)/(dx)(x + y) = (d)/(dx)(y^2) x + y^2 (d)/(dx)( x) (x + y) (1 + (dy)/(dx)) = 2y (dy)/(dx) x + y^2 (- x) (x + y) + (x + y) (dy)/(dx) = 2y x (dy)/(dx) - y^2 x Step 2: Rearrange the equation to group terms with (dy)/(dx) on one side. (x + y) (dy)/(dx) - 2y x (dy)/(dx) = -y^2 x - (x + y) Step 3: Factor out (dy)/(dx). (dy)/(dx)((x + y) - 2y x) = -y^2 x - (x + y) Step 4: Solve for (dy)/(dx). (dy)/(dx) = (-y^2 x - (x + y))/((x + y) - 2y x) (dy)/(dx) = (y^2 x + (x + y))/(2y x - (x + y)) --- 7. f(x) = ((2x)/(1 - x^2)) Step 1: Recall the derivative of (u) is (1)/(1+u^2) (du)/(dx). Let u = (2x)/(1 - x^2). Step 2: Find (du)/(dx) using the quotient rule. Let p = 2x and q = 1 - x^2. Then p' = 2 and q' = -2x. (du)/(dx) = (p'q - pq')/(q^2) = ((2)(1 - x^2) - (2x)(-2x))/((1 - x^2)^2) (du)/(dx) = (2 - 2x^2 + 4x^2)/((1 - x^2)^2) = (2 + 2x^2)/((1 - x^2)^2) = (2(1 + x^2))/((1 - x^2)^2) Step 3: Substitute u and (du)/(dx) into the derivative formula for (u). f'(x) = (1)/(1 + (2x)1 - x^2)^2 · (2(1 + x^2))/((1 - x^2)^2) f'(x) = (1)/(1 + 4x^2)(1 - x^2)^2 · (2(1 + x^2))/((1 - x^2)^2) f'(x) = (1)/((1 - x^2)^2 + 4x^2)(1 - x^2)^2 · (2(1 + x^2))/((1 - x^2)^2) f'(x) = ((1 - x^2)^2)/((1 - x^2)^2 + 4x^2) · (2(1 + x^2))/((1 - x^2)^2) f'(x) = (2(1 + x^2))/((1 - x^2)^2 + 4x^2) Step 4: Expand the denominator. (1 - x^2)^2 + 4x^2 = 1 - 2x^2 + x^4 + 4x^2 = 1 + 2x^2 + x^4 = (1 + x^2)^2 Step 5: Substitute the simplified denominator back into f'(x). f'(x) = (2(1 + x^2))/((1 + x^2)^2) f'(x) = (2)/(1 + x^2) Note: This result is consistent with the identity 2(x) = ((2x)/(1-x^2)) for |x|<1, as (d)/(dx)(2(x)) = (2)/(1+x^2). --- 8. y = ( x + x) Step 1: Use the identity x + x = e^x. y = (e^x) Step 2: Simplify the function using logarithm properties. y = x Step 3: Differentiate y with respect to x. (dy)/(dx) = (d)/(dx)(x) (dy)/(dx) = 1 --- 9. y = e^e^x Step 1: Apply the chain rule. The derivative of e^u is e^u (du)/(dx). Let u = e^x. (dy)/(dx) = e^u · (du)/(dx) Step 2: Find (du)/(dx). (du)/(dx) = (d)/(dx)(e^x) = e^x Step 3: Substitute u and (du)/(dx) back into the expression for (dy)/(dx). (dy)/(dx) = e^e^x · e^x (dy)/(dx) = e^x e^e^x --- 10. y = (x)/(x^2 + 1) (Find (d^2y)/(dx^2)) Step 1: Find the first derivative (dy)/(dx) using the quotient rule. Let u = x and v = x^2 + 1. Then u' = 1 and v' = 2x. (dy)/(dx) = (u'v - uv')/(v^2) = ((1)(x^2 + 1) - (x)(2x))/((x^2 + 1)^2) (dy)/(dx) = (x^2 + 1 - 2x^2)/((x^2 + 1)^2) (dy)/(dx) = (1 - x^2)/((x^2 + 1)^2) Step 2: Find the second derivative (d^2y)/(dx^2) by differentiating (dy)/(dx) using the quotient rule again. Let p = 1 - x^2 and q = (x^2 + 1)^2. Then p' = -2x. For q', use the chain rule: q' = 2(x^2 + 1) · (2x) = 4x(x^2 + 1). (d^2y)/(dx^2) = (p'q - pq')/(q^2) (d^2y)/(dx^2) = ((-2x)(x^2 + 1)^2 - (1 - x^2)(4x(x^2 + 1)))/(((x^2 + 1)^2)^2) (d^2y)/(dx^2) = ((-2x)(x^2 + 1)^2 - 4x(1 - x^2)(x^2 + 1))/((x^2 + 1)^4) Step 3: Factor out common terms from the numerator, specifically -2x(x^2 + 1). (d^2y)/(dx^2) = (-2x(x^2 + 1)[(x^2 + 1) + 2(1 - x^2)])/((x^2 + 1)^4) (d^2y)/(dx^2) = (-2x[(x^2 + 1) + 2 - 2x^2])/((x^2 + 1)^3) (d^2y)/(dx^2) = (-2x[3 - x^2])/((x^2 + 1)^3) (d^2y)/(dx^2) = (2x(x^2 - 3))/((x^2 + 1)^3) That's 2 down. 3 left today — send the next one.