Find the second derivative of the function \( F(t)=5 t \sin \left(t^{5}\right) \)

To find the second derivative of the function \( F(t)=5 t \sin \left(t^{5}\right) \), we will first compute the first derivative \( F'(t) \) using the product rule, and then differentiate again to find \( F''(t) \). 1. **Find the first derivative \( F'(t) \)**: We can use the product rule where \( u = 5t \) and \( v = \sin(t^5) \). The product rule states \( (uv)' = u'v + uv' \). - \( u' = 5 \) - To find \( v' \), we need the chain rule: \[ v' = \cos(t^5) \cdot \frac{d}{dt}(t^5) = \cos(t^5) \cdot 5t^4 \] Now applying the product rule: \[ F'(t) = u'v + uv' = 5 \sin(t^5) + 5t \cdot 5t^4 \cos(t^5) \] Simplifying, we have: \[ F'(t) = 5 \sin(t^5) + 25t^5 \cos(t^5) \] 2. **Find the second derivative \( F''(t) \)**: Now, we differentiate \( F'(t) = 5 \sin(t^5) + 25t^5 \cos(t^5) \). - The derivative of the first term: \[ \frac{d}{dt}(5 \sin(t^5)) = 5 \cos(t^5) \cdot 5t^4 = 25t^4 \cos(t^5) \] - For the second term \( 25t^5 \cos(t^5) \), we again use the product rule: - Let \( a = 25t^5 \) and \( b = \cos(t^5) \). - \( a' = 125t^4 \) - \( b' = -\sin(t^5) \cdot 5t^4 = -5t^4 \sin(t^5) \) Thus, \[ \frac{d}{dt}(25t^5 \cos(t^5)) = a'b + ab' = 125t^4 \cos(t^5) + 25t^5(-5t^4 \sin(t^5)) \] Simplifying: \[ = 125t^4 \cos(t^5) - 125t^9 \sin(t^5) \] Now, we can add both derivatives for \( F''(t) \): \[ F''(t) = 25t^4 \cos(t^5) + (125t^4 \cos(t^5) - 125t^9 \sin(t^5)) \] Combining like terms: \[ F''(t) = (25t^4 + 125t^4) \cos(t^5) - 125t^9 \sin(t^5) \] \[ = 150t^4 \cos(t^5) - 125t^9 \sin(t^5) \] Thus, the second derivative of the function is: \[ F''(t) = 150t^4 \cos(t^5) - 125t^9 \sin(t^5) \]