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The Product Rule is a fundamental tool used in calculus to find the derivative of the product of two functions. In simple terms, if you have a function that is the product of two separate functions, such as \(f(x) = u(x) \cdot v(x)\), the Product Rule helps you find the derivative \(f'(x)\). The rule is expressed as:\[f'(x) = u'(x) v(x) + u(x) v'(x)\]This means you need to do two things:

• Take the derivative of the first function \(u(x)\), and multiply it by the second function \(v(x)\).
• Then, take the derivative of the second function \(v(x)\), and multiply it by the first function \(u(x)\).

Finally, sum these two products together to find the derivative of the whole function.

In our example, we identified \(u(x) = 2x - 3\) and \(v(x) = x^{3/2}\). Using the Product Rule, we find the derivative as\[f'(x) = (2)(x^{3/2}) + (2x - 3)(\frac{3}{2}x^{1/2})\].Simplifying these terms gives us the final derivative of the function.

The Power Rule is one of the easiest and most straightforward rules in differentiation. It is a quick method for finding derivatives of functions where the variable \(x\) is raised to some power. The Power Rule states:

If a function is in the form \(x^n\), then the derivative \(f'(x)\) is \(nx^{n-1}\).

For example, let's consider the function \(v(x) = x^{3/2}\). To apply the Power Rule, take the exponent \(3/2\) in front of \(x\) and subtract one from the exponent. This gives:\[v'(x) = \frac{3}{2}x^{(3/2)-1} = \frac{3}{2}x^{1/2}\]Make sure you carefully calculate the new exponent when applying the Power Rule, especially with fractions. In our exercise solution, the Power Rule was essential for finding the derivative of \(v(x)\), so we could correctly apply the Product Rule.

Derivatives are a core concept of calculus that measure how a function changes as its input changes. They are the foundation for understanding rates of change and slopes of curves.

If you have a function \(f(x)\), its derivative \(f'(x)\) gives you the rate at which \(f(x)\) changes with respect to \(x\). You can think of this as the slope of the tangent line to the curve at any point.

In solving our exercise, the derivatives \(u'(x) = 2\) and \(v'(x) = \frac{3}{2}x^{1/2}\) tell us how each part of the function \((2x-3)\) and \(x^{3/2}\), respectively, change. These derivatives are crucial for applying the Product Rule and determining \(f'(x)\), the derivative of the entire function. Understanding derivatives helps us not only find the slope but also optimize functions and predict future results in real-world applications.