91Ó°ÊÓ

In calculus, the product rule is essential when differentiating functions that are multiplied together. If you have two functions, say \( u(x) \) and \( v(x) \), that are multiplied, you can't simply differentiate them separately and multiply the results. Instead, the product rule is applied.

• The product rule formula is given by \( (uv)' = u'v + uv' \).
• This means that you differentiate the first function \( u(x) \) and multiply it by the second function \( v(x) \), and then add it to the product of the first function \( u(x) \) and the derivative of the second function \( v(x) \).

It’s handy to write down which part of the function corresponds to \( u(x) \) and which part corresponds to \( v(x) \) before applying the product rule, as seen in the example where \( u(x) = x^4 \) and \( v(x) = e^x \). The derivatives, \( u'(x) = 4x^3 \) and \( v'(x) = e^x \), are then used to find the derivative of the product \( f(x) = x^4 e^x \).
This methodical approach helps avoid mistakes and ensures accurate results.

The first derivative of a function provides the rate at which the function value is changing with respect to changes in the input variable. In simpler terms, it gives the slope of the function at any given point.

• For the function \( f(x) = x^4 e^x \), we initially determine its first derivative using the product rule.
• After computing the derivatives for \( u(x) = x^4 \) as \( 4x^3 \) and \( v(x)= e^x \) as \( e^x \), we applied the product rule yielding \( f'(x) = e^x (4x^3 + x^4) \).

This result implies that the function \( f(x) \) grows at a rate proportional to \( e^x (4x^3 + x^4) \), combining constant growth due to the exponential factor and polynomial growth through the \( x^4 \) term. Understanding the behavior of this derivative lets you explore how the shape of the function evolves, identifying regions of increase or decrease.

The second derivative is all about the rate of change of the rate of change. It's a powerful tool for examining the concavity of a function and helps to identify points of inflection where the function changes its curvature.

• For a complex function like \( f(x) = x^4 e^x \), finding the second derivative involves differentiating the first derivative \( f'(x) = e^x (4x^3 + x^4) \) again.
• This step uses the product rule once more, identifying \( a(x) = e^x \) and \( b(x) = 4x^3 + x^4 \).
• Calculate \( a'(x) = e^x \) and \( b'(x) = 12x^2 + 4x^3 \), then apply the product rule for a detailed second derivative: \( f''(x) = e^x (16x^3 + x^4 + 12x^2) \).

Analyzing the second derivative provides insights into the function's acceleration or deceleration. You can also pinpoint where it changes from concave up to concave down, offering valuable information about the function's overall behavior.