Problem 43 - Calculus - AP Edition

Question

Find the differential \(d y\). (a) \(y=4 x^{3}-7 x^{2}\) (b) \(y=x \cos x\)

Step-by-Step Solution

Answer

(a) \( dy = (12x^2 - 14x) \, dx \); (b) \( dy = (\cos x - x \sin x) \, dx \).

Step 1: Identify the function (a)

For (a), the function is given as \( y = 4x^3 - 7x^2 \). We are tasked with finding the differential \( dy \).

Step 2: Differentiate the function (a)

To find \( dy \), differentiate \( y \) with respect to \( x \).The derivative of \( y = 4x^3 \) is \( 12x^2 \) and the derivative of \( -7x^2 \) is \( -14x \). So, the derivative \( \frac{dy}{dx} = 12x^2 - 14x \).

Step 3: Express the differential (a)

The differential \( dy \) is obtained by multiplying the derivative by \( dx \):\[ dy = (12x^2 - 14x) \, dx \]

Step 4: Identify the function (b)

For (b), the function is given as \( y = x \cos x \). Again, we are tasked with finding the differential \( dy \).

Step 5: Differentiate the function (b)

To differentiate \( y = x \cos x \), we use the product rule. The product rule states that \((uv)' = u'v + uv'\), where \( u = x \) and \( v = \cos x \).- \( u' = 1 \) and \( v' = -\sin x \).Thus, \( \frac{dy}{dx} = 1 \cdot \cos x + x (-\sin x) = \cos x - x \sin x \).

Step 6: Express the differential (b)

The differential \( dy \) for function (b) is:\[ dy = (\cos x - x \sin x) \, dx \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, representing the process of finding the derivative of a function. It measures how a function changes as its input changes, offering a way to "differentiate" or "distinguish" its rate of change. In simple terms, differentiation involves calculating the slope of a tangent line to the curve of a function at any given point. Here's why differentiation is crucial:

• It helps in determining rates of change, such as velocity or acceleration in physics.
• It's used to find local maxima or minima, revealing where functions increase or decrease.
• It provides insights into the behavior of functions, aiding in the study of curves, motion, and other phenomena.

Differentiation follows rules that enable us to break down complex expressions into simpler components. For instance, in problem (a) from the exercise, we dealt with a polynomial function, which is straightforward to differentiate. By understanding the rules, like power and constant rules, the process becomes systematic and predictable.

Product Rule

When dealing with the differentiation of products of two functions, the Product Rule is key. It helps us in finding the derivative of a function that is the product of two functions. The Product Rule states:\[(uv)' = u'v + uv'\]where \(u\) and \(v\) are two differentiable functions.In exercise (b), we faced an example where the Product Rule was essential. The function given was \(y = x \cos x\), involving a multiplication of \(x\) with \(\cos x\). To differentiate it, we set \(u = x\) and \(v = \cos x\). Using the Product Rule:

• Calculate \(u' = 1\) for the function \(u = x\) since the derivative of \(x\) is 1.
• Find \(v' = -\sin x\) for \(v = \cos x\), since the derivative of \(\cos x\) is \(-\sin x\).
• Apply the rule: \(\frac{dy}{dx} = 1 \cdot \cos x + x(-\sin x) = \cos x - x\sin x\)

By understanding and applying the Product Rule, complex functions that aren't mere sums or differences become much easier to tackle in differentiation tasks.

Derivatives

Derivatives are the result of differentiation, representing the function that describes the rate of change of another function. For any function \(y = f(x)\), its derivative \(\frac{dy}{dx}\) signifies how \(y\) changes with respect to \(x\). Here's the lowdown on derivatives:

• They serve as the backbone of differential calculus, appearing in countless applications from engineering to economics.
• A derivative can be seen as a linear approximation of a function near a given point.
• They help identify the nature of stationary points, like turning points and inflection points.

In our exercises, we explored derivatives of polynomial and trigonometric functions. In part (a), after differentiating \(y = 4x^3 - 7x^2\), the derivative \(\frac{dy}{dx} = 12x^2 - 14x\) showed us how the original polynomial behaves for different \(x\) values. Similarly, part (b) reveals \(\frac{dy}{dx} = \cos x - x\sin x\), offering insights on how the product function changes along its domain. Understanding derivatives not only helps with immediate calculations but also builds a deeper intuition about the changing nature of functional relationships.