91Ó°ÊÓ

When dealing with the differentiation of products of functions, the product rule is essential. The product rule allows us to differentiate two functions that are multiplied together. In our given function \( v(t) = (1-t)(1+t^2)^{-1} \), we can identify two separate functions: \( u(t) = 1-t \) and \( w(t) = (1+t^2)^{-1} \).

The product rule formula states that if you have functions \( u \) and \( w \), their product \( y = u \times w \) will have a derivative given by:

\[ \frac{dy}{dt} = u'w + uw' \]

Here, \( u' \) and \( w' \) represent the derivatives of \( u \) and \( w \) respectively. To apply the product rule, differentiate each function separately, and then substitute into the formula. This approach helps simplify the differentiation process when dealing with multiplied functions.

The chain rule is a fundamental method in calculus used to differentiate composite functions. In our problem, the function \( w(t) = (1+t^2)^{-1} \) is a perfect candidate for using the chain rule. This function can be decomposed into simpler parts where an outer function and an inner function can be identified.

Firstly, let's rewrite \( w(t) \) as \( y = x^{-1} \), where \( x = 1+t^2 \). Here, \( x^{-1} \) is the outer function and \( 1+t^2 \) is the inner function. To find the derivative \( w'(t) \), the chain rule suggests that we take the derivative of the outer function with respect to the inner function, and multiply it by the derivative of the inner function itself.

The derivative of \( x^{-1} \) with respect to \( x \) is \( -x^{-2} \). Multiply this by the derivative of the inner function \( 1+t^2 \) which is \( 2t \). Putting it all together, we get:

\[ w'(t) = -2t(1+t^2)^{-2} \]

This application of the chain rule allows us to handle derivatives of functions that involve compositions of other functions smoothly.

Differentiation is a core concept in calculus that deals with finding the rate at which a quantity changes. In our specific problem, we aim to find the derivative of \( v(t) = (1-t)(1+t^2)^{-1} \).

To achieve this, we broke down the function using both the product rule and the chain rule. Differentiation involves the following:

• Identifying components of the function (like \( u(t) \) and \( w(t) \) in our example).
• Finding the derivatives of each component separately using known rules.
• Combining these derivatives according to the product rule.

The final step after using the differentiation process is to simplify. In the exercise, after applying these rules, the derivative simplifies to:

\[ \frac{dv}{dt} = -(1+t^2)^{-1} + (-2t + 2t^2)(1+t^2)^{-2} \]

This expression fully represents the rate of change of the original function \( v(t) \) with respect to \( t \). Understanding differentiation allows you to find how outputs change with respect to inputs, which is crucial in analyzing functions and real-world applications.