91Ó°ÊÓ

The product rule is a technique used in calculus to differentiate a product of two functions. Suppose you have two functions, say, \(u(x)\) and \(v(x)\). The product rule states that the derivative of their product is given by:

\[ (uv)' = u'v + uv'. \]

To apply this, let's take a look at an example from our exercise. Suppose we need to differentiate \( y = x (x-3)^4 \). Here, we identify \( u = x \) and \( v = (x-3)^4 \).

First, differentiate \( u \) and \( v \) separately:

• \( u' = 1 \)
• Next, for \( v \), we use the chain rule to find that: \[ v' = 4(x-3)^3 \cdot 1 = 4(x-3)^3 \]

Applying the product rule, we get:
\( y' = u'v + uv' = 1 \times (x-3)^4 + x \times 4(x-3)^3 \),
which simplifies to: \( y' = (x-3)^4 + 4x(x-3)^3 \).
For further simplification: \( y' = (x-3)^3 ((x-3) + 4x) = (x-3)^3 (5x-3) \).

The chain rule is a fundamental method in calculus for differentiating compositions of functions. If you have a function \( y = g(f(x)) \), the chain rule tells us how to find \( y' \):

\[ y' = g'(f(x)) \times f'(x). \]

Let's understand this with an example from our exercise where we need to differentiate \( y = x \sqrt{2x-3} \). To use the product rule here, set \( u = x \) and \( v = (2x-3)^{1/2} \).

First, differentiate \( u \) and \( v \) separately:

• \( u' = 1 \)
• Using the chain rule on \( v = (2x-3)^{1/2} \), let \( f(x) = 2x-3 \) and \( g(t) = t^{1/2} \). Then, \( v = g(f(x)) \), and we find \( v' \) as follows: \[ v' = g'(f(x)) \times f'(x) = \frac{1}{2}(2x-3)^{-1/2} \times 2 = \frac{1}{\sqrt{(2x-3)}} \]

Applying the product rule, we get: \( y' = u'v + uv' \). Substituting, we find: \( y' = 1 \sqrt{2x-3} + x \frac{1}{\sqrt{2x-3}} \).

Simplifying, we get: \( y' = \frac{(2x-3) + x}{\sqrt{2x-3}} \), which can be further simplified to: \( y' = \frac{3(x-1)}{\sqrt{2x-3}} \).

The quotient rule is essential for differentiating a quotient of two functions. When you have \( y = \frac{u}{v} \), the quotient rule states:

\[ (\frac{u}{v})' = \frac{u'v - uv'}{v^2}. \]

Let's start with an example from our exercise: differentiate \( y = \frac{x}{x+5} \). Identify \( u = x \) and \( v = x+5 \).

• Compute \( u' = 1 \)
• For \( v \), \( v' = 1 \).

Applying the quotient rule, we find:
\( y' = \frac{1(x+5) - x(1)}{(x+5)^2} \) \( y' = \frac{5}{(x+5)^2} \).

As another example, consider differentiating \( y = \frac{x}{x^2+1} \). Here, \( u = x \) and \( v = x^2+1 \).

• Compute \( u' = 1 \)
• For \( v \), \( v' = 2x \).

Using the quotient rule, we get:
\( y' = \frac{1(x^2 + 1) - x(2x)}{(x^2 + 1)^2} \), which simplifies to:

\( y' = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} \), and further to: \( y' = \frac{1-x^2}{(x^2 + 1)^2} \).