91Ó°ÊÓ

When differentiating a function that is a ratio of two other functions, we use the Quotient Rule. This rule helps find the derivative of a quotient. It is given by the formula: \[ \frac{d}{dx} \bigg( \frac{u}{v} \bigg) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \] where:

• u is the numerator function
• v is the denominator function

In our problem, the numerator is \( x + 3 \) and the denominator is \( (2x + 1)^2 \). Begin by identifying these components so we can apply the rule smoothly. Ensure you differentiate both the numerator and the denominator independently before plugging them back into the Quotient Rule formula.

The Chain Rule is crucial when differentiating compositions of functions. In essence, it helps us find the derivative of a function nested inside another function. The Chain Rule states:\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x) \]In this exercise, the denominator \( (2x + 1)^2 \) is a composition of functions where:

• The outer function is \( y = w^2 \)
• The inner function is \( w = 2x + 1 \)

First, let’s differentiate the outer function \( w^2 \) with respect to \( w \), which gives \( 2w \). Next, differentiate the inner function \( 2x + 1 \) with respect to \( x \), which is \( 2 \). Combining these using the Chain Rule, we get:\[ \frac{dv}{dx} = 2(2x + 1) \times 2 = 4(2x + 1) \]

Simplification is the last step in solving differentiation problems. Applying the Quotient Rule can sometimes yield a messy expression, so we need to tidy it up. From our previously identified parts, the numerator after applying the Quotient Rule is obtained as: \[ (2x+1)^2 - (x+3) \times 4(2x+1) = 4x^2 + 4x + 1 - 4(x + 3)(2x+1) \]Expanding and simplifying each part systematically: \[ 4x^2 + 4x + 1 - 4x(2x+1) - 12(2x+1) = 4x^2 + 4x + 1 - 8x^2 - 4x - 24 \]Combining like terms, the expression simplifies to: \[ -4x^2 - 23 \]Finally, placing this over the denominator squared, we get the simplified derivative: \[ \frac{-4x^2 - 23}{(2x+1)^4} \]By breaking down each term properly, the final simplified derivative is much easier to work with and understand.