91Ó°ÊÓ

When you encounter a function that is the product of two other functions, you will need to use the product rule for differentiation. This is a crucial concept in calculus and is essential for solving many problems. The product rule allows you to find the derivative of a product of two functions. It states that if you have two functions, let's call them \( f(t) \) and \( g(t) \), and their product is \( v(t) = f(t) \times g(t) \), then the derivative of \( v(t) \) is given by:
\( v'(t) = f'(t)g(t) + f(t)g'(t) \)
In other words, you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.
In our problem, we identified:
- \( f(t) = t^5 \) and \( g(t) = e^{-ct} \)
The product rule helps us by breaking the problem into manageable parts.

The chain rule is used to differentiate composite functions. In simpler terms, when you have a function inside another function, the chain rule is your go-to method. The chain rule states that if you have a composite function \( h(x) = f(g(x)) \), its derivative is:
\( h'(x) = f'(g(x)) \times g'(x) \)
Essentially, you differentiate the outer function and multiply it by the derivative of the inner function.
In our exercise, we used the chain rule to differentiate \( g(t) = e^{-ct} \). Here, the outer function is the exponential function, and the inner function is \( -ct \).
The derivative of the exponential function \( e^{u} \) is \( e^u \). Applying the chain rule, we get:
\( \frac{d}{dt}e^{-ct} = e^{-ct} \times (-c) = -ce^{-ct} \)
This derivative is crucial for applying the product rule.

Let's bring it all together using the product rule and the chain rule to compute the derivative of the given function. Our original function is \( v(t) = t^5 e^{-ct} \). First, we find the derivatives of the individual parts:
- \( f(t) = t^5 \) so \( f'(t) = 5t^4 \)
- \( g(t) = e^{-ct} \) and using the chain rule, we find \( g'(t) = -ce^{-ct} \)
Next, we apply the product rule:
\[ v'(t) = f'(t)g(t) + f(t)g'(t) \]
Substitute the derivatives we found into the formula:
\[ v'(t) = 5t^4 e^{-ct} + t^5(-ce^{-ct}) \]
Finally, we simplify the expression:
\[ v'(t) = 5t^4 e^{-ct} - ct^5 e^{-ct} = t^4 e^{-ct} (5 - ct) \]
So, the derivative of \( v(t) = t^5 e^{-ct} \) is:
\[ v'(t) = t^4 e^{-ct} (5 - ct) \].
Breaking it all down like this makes it easier to see how we use the product rule and chain rule effectively.