Find the second derivative of the function \( F(t)=6 t \sin \left(t^{6}\right) \)

To find the second derivative of the function \( F(t) = 6t \sin(t^6) \), we start by finding the first derivative \( F'(t) \). Using the product rule for differentiation, where \( u = 6t \) and \( v = \sin(t^6) \), we have: \[ F'(t) = u'v + uv' \] Calculating \( u' \) and \( v' \): 1. \( u' = 6 \) 2. \( v = \sin(t^6) \) and using the chain rule for its derivative: \[ v' = \cos(t^6) \cdot (t^6)' = \cos(t^6) \cdot 6t^5 \] Now plug back into the product rule: \[ F'(t) = 6 \sin(t^6) + 6t \cdot \cos(t^6) \cdot 6t^5 \] This simplifies to: \[ F'(t) = 6 \sin(t^6) + 36t^6 \cos(t^6) \] Now, we find the second derivative \( F''(t) \): Using the derivative of \( F'(t) \): 1. Differentiate \( 6 \sin(t^6) \): \[ (6 \sin(t^6))' = 6 \cos(t^6) \cdot 6t^5 = 36t^5 \cos(t^6) \] 2. Differentiate \( 36t^6 \cos(t^6) \) using the product rule again. Let \( a = 36t^6 \) and \( b = \cos(t^6) \): \[ F''(t) = (36t^6)' \cos(t^6) + 36t^6 (\cos(t^6))' \] Calculating those parts: - \( (36t^6)' = 216t^5 \) - \( (\cos(t^6))' = -\sin(t^6) \cdot 6t^5 \) Now substitute these values back into our expression: \[ F''(t) = 216t^5 \cos(t^6) + 36t^6 (-6t^5 \sin(t^6)) \] This simplifies to: \[ F''(t) = 216t^5 \cos(t^6) - 216t^{11} \sin(t^6) \] Thus, the second derivative of the function \( F(t) \) is: \[ F''(t) = 216t^5 \cos(t^6) - 216t^{11} \sin(t^6) \]