y = (t2 sin t) / (cos t). 2) If y = (t2 sin t) / (cos t) show that t2(d2y/dt2) + 4t(dy/dt) - (t2+6)y = 0. 3) If y = t^n show that dy/dt = nt^(n-1).

1) Find (dy)/(dt) for y = (t^2 t)/( t) Step 1: Rewrite the expression for y using trigonometric identities. We know that ( t)/( t) = t. So, we can write y as: y = t^2 t Step 2: Apply the product rule for differentiation. The product rule states that if y = u · v, then (dy)/(dt) = u'v + uv'. Let u = t^2 and v = t. Then, u' = (d)/(dt)(t^2) = 2t. And v' = (d)/(dt)( t) = ^2 t. Step 3: Substitute u, v, u', v' into the product rule formula. (dy)/(dt) = (2t)( t) + (t^2)(^2 t) (dy)/(dt) = 2t t + t^2 ^2 t The derivative is 2t t + t^2 ^2 t. 2) If y = (t^2 t)/( t), show that t^2 (d^2y)/(dt^2) + 4t (dy)/(dt) - (t^2+6)y = 0. Step 1: Use the result from problem 1 for (dy)/(dt) and the expression for y. We have y = t^2 t and (dy)/(dt) = 2t t + t^2 ^2 t. Step 2: Find the second derivative, (d^2y)/(dt^2). We need to differentiate (dy)/(dt) with respect to t: (d^2y)/(dt^2) = (d)/(dt)(2t t + t^2 ^2 t) Differentiate 2t t using the product rule: (d)/(dt)(2t t) = (2)( t) + (2t)(^2 t) = 2 t + 2t ^2 t. Differentiate t^2 ^2 t using the product rule and chain rule: (d)/(dt)(t^2 ^2 t) = (2t)(^2 t) + (t^2)(2 t · t t) = 2t ^2 t + 2t^2 ^2 t t. Combine these two results for (d^2y)/(dt^2): (d^2y)/(dt^2) = (2 t + 2t ^2 t) + (2t ^2 t + 2t^2 ^2 t t) (d^2y)/(dt^2) = 2 t + 4t ^2 t + 2t^2 ^2 t t Step 3: Substitute y, (dy)/(dt), and (d^2y)/(dt^2) into the left-hand side of the given equation. The left-hand side (LHS) is t^2 (d^2y)/(dt^2) + 4t (dy)/(dt) - (t^2+6)y. Substitute the expressions: LHS = t^2 (2 t + 4t ^2 t + 2t^2 ^2 t t) + 4t (2t t + t^2 ^2 t) (t^2+6)(t^2 t) Step 4: Expand and simplify the expression. Expand each term: t^2 (d^2y)/(dt^2) = 2t^2 t + 4t^3 ^2 t + 2t^4 ^2 t t 4t (dy)/(dt) = 8t^2 t + 4t^3 ^2 t -(t^2+6)y = -t^4 t - 6t^2 t Now, sum these expanded terms: LHS = (2t^2 t + 4t^3 ^2 t + 2t^4 ^2 t t) + (8t^2 t + 4t^3 ^2 t) + (-t^4 t - 6t^2 t) Group terms by t: (2t^2 + 8t^2 - 6t^2 - t^4) t = (4t^2 - t^4) t Group terms by ^2 t: (4t^3 + 4t^3) ^2 t = 8t^3 ^2 t Group terms by ^2 t t: 2t^4 ^2 t t Combining these, the LHS simplifies to: LHS = (4t^2 - t^4) t + 8t^3 ^2 t + 2t^4 ^2 t t This expression does not simplify to 0. It is possible there is a typo in the question. 3) If y = t^n show that (dy)/(dt) = nt^n-1. Step 1: Apply the power rule for differentiation. The power rule states that if y = t^n, then (dy)/(dt) = n t^n-1. Given y = t^n. (dy)/(dt) = n t^n-1 This is a direct application of the power rule. The derivative is nt^n-1. Send me the next one 📸