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The product rule is a key technique in calculus used for differentiating products of two or more functions. It's crucial for functions that aren't simply linear. The product rule states that if you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:\[(uv)' = u'v + uv'\]This rule can be extended to more than two functions, which is useful in our exercise. Here, the function consists of three components: an exponential function \( e^x \), and two polynomial forms \( (z-1) \) and \( (z^2+1) \). To find the first derivative with respect to \( x \), we're treating \( z \) terms as constants. By applying the product rule on two parts step-by-step, we determine how each component contributes to the derivative. This multi-component nature is what makes the product rule so invaluable for such problems in calculus.

Exponential functions are another key aspect of our given problem. In general, an exponential function can be expressed as \( a^x \) where \( a \) is a constant. The most common base used in calculus is the natural base \( e \), and the function becomes \( e^x \). One of the most interesting properties of exponential functions is that they are their own derivatives.

• This means if you differentiate \( e^x \), you get \( e^x \) again!
• This property simplifies working with chains of functions, as seen in our problem where \( e^x \) is the primary function differentiated.

Understanding this helps us easily apply the product rule as the exponential neither complicates nor changes the inner workings or results during differentiation.

The primary goal here is to calculate the first and second derivatives of a function. The first derivative (\( f'(x) \)) gives the rate of change of the function, essentially its slope at any point. For our function \( w = e^x (z-1)(z^2+1) \), the first derivative was straightforward thanks to the product rule and the constant nature of \( z \) with respect to \( x \).

• The first derivative \( \frac{dw}{dx} = e^x (z-1)(z^2+1) \) remains simple because the internal derivatives of \( z \) terms equal zero.
• The second derivative \( \frac{d^2w}{dx^2} = 2e^x (z-1)(z^2+1) \) involves applying the derivative operation once more to our first result.

This second derivative expands our understanding by showing how things like acceleration or curvature behave, giving a richer interpretation of the function’s behavior beyond mere slopes. Understanding these concepts allows better insights into complex mathematical and real-world phenomena.