Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\cos \left(\frac{3 x^{2}}{x+2}\right) $$

Step-by-Step Solution

Answer

\(D_{x} y = -\frac{3x^2 + 12x}{(x+2)^2} \cdot \sin\left(\frac{3x^2}{x+2}\right)\).

Step 1: Identify the Outer Function

The outer function in this problem is the cosine function. We can represent it as \(y = ext{cos}(u)\), where \(u = \frac{3x^2}{x+2}\). Our task first involves taking the derivative of the cosine function with respect to \(u\).

Step 2: Differentiate the Outer Function

Using the derivative of the cosine function, \(\frac{d}{du}(\cos u) = -\sin u\), find the derivative of the outer function with respect to \(u\). This gives us \(\frac{d}{du}\left(\cos\left(\frac{3x^2}{x+2}\right)\right) = -\sin\left(\frac{3x^2}{x+2}\right)\).

Step 3: Identify the Inner Function

Now we focus on the inner function, \(u = \frac{3x^2}{x+2}\). We need to differentiate this with respect to \(x\).

Step 4: Use the Quotient Rule on the Inner Function

The inner function is a quotient, so apply the quotient rule: \(\frac{d}{dx}\left(\frac{3x^2}{x+2}\right) = \frac{(x+2)\cdot 6x - 3x^2 \cdot 1}{(x+2)^2}\). Simplify this to obtain \(\frac{6x(x+2) - 3x^2}{(x+2)^2} = \frac{6x^2 + 12x - 3x^2}{(x+2)^2} = \frac{3x^2 + 12x}{(x+2)^2}\).

Step 5: Combine Using the Chain Rule

Now, use the chain rule to find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{d}{du}(\cos u) \cdot \frac{du}{dx}\). Substitute the derivatives found: \(\frac{dy}{dx} = -\sin\left(\frac{3x^2}{x+2}\right) \cdot \frac{3x^2 + 12x}{(x+2)^2}\).

Step 6: Final Expression for Derivative

Combine all the parts to express the derivative compactly: \(D_{x} y = -\frac{3x^2 + 12x}{(x+2)^2} \cdot \sin\left(\frac{3x^2}{x+2}\right)\).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions, or functions within functions. When a function is nested inside another, like in the exercise where \(y = \cos(u)\) and \(u = \frac{3x^2}{x+2}\), the chain rule helps us differentiate these layers one at a time.
To apply the chain rule, we differentiate the outer function first, considering the inside function as a single unit (in this case, \(u\)). Then we multiply by the derivative of the inner function.

• Outer Function: \(y = \cos(u)\); its derivative is \(-\sin(u)\).
• Inner Function: \(u = \frac{3x^2}{x+2}\); we use the quotient rule to differentiate.

After obtaining both derivatives, we multiply them together: \( \frac{dy}{dx} = \frac{d}{du}(\cos u) \cdot \frac{du}{dx} \). This gives us the final derivative for the original composite function.

Quotient Rule

The quotient rule is essential for differentiating functions that are divided by each other, such as \(\frac{3x^2}{x+2}\) from the exercise.
It provides a method for finding the derivative of the ratio of two functions.
Here's how it works:
If you have a function \( \frac{v(x)}{w(x)}\), the quotient rule states that its derivative is given by:\[\frac{d}{dx}\left(\frac{v(x)}{w(x)}\right) = \frac{w(x)\cdot v'(x) - v(x)\cdot w'(x)}{(w(x))^2}\].

• Numerator \(v(x)\): \(3x^2\), with derivative \(6x\).
• Denominator \(w(x)\): \(x+2\), with derivative 1.

Substituting these derivatives into the quotient rule formula provides the derivative for the inner function, which is then used in the chain rule calculation for the entire problem.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, appear frequently in calculus. Their derivatives are key to solving many differentiation problems. In our exercise, the outer function is \(\cos(u)\).
Understanding how to differentiate these functions is crucial.

• Derivative of Cosine: The derivative of \(\cos(u)\) is \(-\sin(u)\). This reflects the periodic and oscillating nature of the cosine curve and its slope variations.
• Application: Use this derivative to handle the outer function in the chain rule.

Knowing these derivatives by heart allows for swift application in composite functions, aiding in the seamless solution of derivative problems.