91Ó°ÊÓ

The product rule is a vital concept in differentiation, especially when dealing with products of two functions. In simpler terms, it helps us find the derivative of two functions that are multiplied together.
For two functions \(u(x)\) and \(v(x)\), the product rule states:

• \((uv)' = u'v + uv'\)

This formula might look a bit intimidating, but the idea is straightforward:
First, you differentiate the first function, then multiply it by the second function (without differentiating it).
Next, you keep the first function as it is, and differentiate the second function, then multiply these.
The sum of these products gives you the derivative of the original function.
In our example, we have \(f(x) = e^{x/2} \cdot \sqrt{x-1}\). By identifying \(u(x) = e^{x/2}\) and \(v(x) = (x-1)^{1/2}\),
we apply the product rule to find the derivative \(f'(x)\). This method simplifies finding derivatives when dealing with multiplicative functions.

The chain rule is a handy tool when you're working with composite functions, meaning a function within another function.
The rule helps us differentiate such functions step by step.
If we have a composite function like \(f(g(x))\), the chain rule states:

• \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \)

This tells us to:

• First, differentiate the outer function \(f\) while keeping the inside \(g(x)\) intact.
• Then, multiply by the derivative of the inside function \(g(x)\).

In our example, to differentiate \(u(x) = e^{x/2}\), we let \(g(x) = \frac{x}{2}\) and recognize \(e^{x/2}\) as the outer function. Differentiating this using the chain rule gives us \(u'(x) = e^{x/2} \cdot \frac{1}{2}\), showing how smoothly the chain rule handles composite functions.

The power rule simplifies the differentiation of functions that involve powers of an expression.
If you encounter an expression like \(x^n\), you can quickly find its derivative using the power rule:

• \( \frac{d}{dx} [x^n] = nx^{n-1} \)

This means we bring the power \(n\) down as a coefficient and reduce the power of \(x\) by one.
In the given problem, to differentiate \(v(x) = (x-1)^{1/2}\), we apply the power rule with the expression \(x-1\) raised to a power of \(1/2\).
By thinking of \(h(x) = x-1\) as the core function and using the power rule, we find \(v'(x) = \frac{1}{2}(x-1)^{-1/2} \cdot 1\).
This showcases how the power rule streamlines finding derivatives, especially for functions involving roots or exponents.