The image you uploaded contains a calculus problem regarding the derivative of a product of functions. The problem is in French and is titled "Dérivée d'un produit," which translates to "Derivative of a product." The task involves calculating the derivative of a function that is the product of two other functions.

Here’s a general breakdown of how to solve such a problem:

Product Rule for Derivatives

If you have two functions, say $u(x)$ and $v(x)$, the derivative of their product is given by the product rule:

$\frac{d}{dx} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

This means that you take the derivative of the first function ($u'(x)$) and multiply it by the second function $v(x)$, then add the product of the first function $u(x)$ and the derivative of the second function $v'(x)$.

Problem Breakdown

1. Identify the two functions $u(x)$ and $v(x)$ from the product given in the problem.
2. Differentiate each function individually.
3. Apply the product rule by multiplying the derivative of the first function by the second, and then adding the product of the first function and the derivative of the second function.

Example

If the problem provides a function $y(x) = x^2 \cdot e^x$, where $u(x) = x^2$ and $v(x) = e^x$:

1. Differentiate $u(x) = x^2$: $u'(x) = 2x$

2. Differentiate $v(x) = e^x$: $v'(x) = e^x$

3. Apply the product rule: $y'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ $y'(x) = 2x \cdot e^x + x^2 \cdot e^x$ $y'(x) = e^x(2x + x^2)$

Please confirm if you want to proceed with solving the specific problem shown in the image or if you need further assistance with the concept.

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Related Questions:

1. What is the derivative of the product of three functions?
2. How does the product rule change if one of the functions is a constant?
3. Can the product rule be applied to vector functions?
4. How would the product rule be used in a real-world application?
5. How do you differentiate a product if one function is implicit?

Tip:

When dealing with products of functions, always check if any simplifications or factorizations are possible before applying the product rule to make the calculations easier.