91Ó°ÊÓ

A vector function is a mathematical expression that includes multiple components, each of which can depend on a variable—in this case, represented by a vector. In the exercise, we were tasked with finding the derivative of the vector function \( f(x) = \left(x^2, x \sin x, 2x + 3\right) \). Each component of this function is differentiable separately. The goal is to take the derivative of each component function with respect to \( x \) and understand how these derivatives contribute to the overall derivative of the vector function.

• The first component \( f_1(x) = x^2 \) is a simple polynomial.
• The second component \( f_2(x) = x \sin x \) involves trigonometry and multiplication.
• The third component \( f_3(x) = 2x + 3 \) is a straightforward linear function.

Ultimately, the derivative of the vector function \( Df(x) \) is another vector: \( \left( Df_1(x), Df_2(x), Df_3(x) \right) \), each entry representing the derivative of the corresponding component.

The product rule is crucial when differentiating products of two functions. It states that if you have a function \( y = u(x) \cdot v(x) \), the derivative \( y' \) is given by \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This rule was applied in the second component of the vector function, \( f_2(x) = x \sin x \). Here, \( u(x) = x \) and \( v(x) = \sin x \).

• Differentiate \( u(x) \): \( u'(x) = 1 \).
• Differentiate \( v(x) \): \( v'(x) = \cos x \).
• Apply the product rule: \( Df_2(x) = 1 \cdot \sin x + x \cdot \cos x \).

This method efficiently allows us to handle derivatives involving trigonometric functions and algebraic expressions multiplied together.

The power rule is a fundamental principle for differentiating functions of the form \( x^n \). According to this rule, the derivative of \( x^n \) is \( n \cdot x^{n-1} \). It's simple yet powerful, offering quick solutions for polynomial expressions. In the first component, \( f_1(x) = x^2 \), the power rule swiftly provides the derivative:

• The exponent \( n \) is 2.
• Apply the power rule: \( Df_1(x) = 2 \cdot x^{2-1} = 2x \).

This makes differentiating polynomial components effortless and is often among the first rules taught in calculus courses due to its importance and simplicity.

The derivative of a linear function, defined as \( ax + b \), is straightforward: it's simply the coefficient \( a \). This is because the slope of a linear function doesn't change, and differential calculus measures how a function's output changes relative to changes in the input.
In the exercise, for the third component \( f_3(x) = 2x + 3 \), the derivative simplifies to:

• Identify the coefficient of \( x \): 2.
• As \( b \) is constant, its derivative is 0.
• The derivative is thus \( Df_3(x) = 2 \).

This method highlights how simple changes in the input of a linear function lead to predictable changes in the output, reinforcing the concept of constant rates of change.