Problem 11 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate. $$ \frac{d}{d x} \sqrt{\sin \left(2 x^{3}\right)} $$

Step-by-Step Solution

Answer

Hence the derivative of \(\sqrt{\sin(2x^3)}\) is \(3x^2 \cdot \frac{\cos(2x^3)}{\sqrt{\sin(2x^3)}}\)

Step 1: Apply Chain Rule

The chain rule is used when the function is composite i.e., function inside function. The given function has a square root function, sine function and a polynomial function. So, let's apply chain rule which states that the derivative of a composite function is the derivative of the outer function multiplied by the derivative of the inner function. Therefore,\[ \frac{d}{dx} \sqrt{\sin(2x^3)} = \frac{1}{2 \sqrt{\sin(2x^3)}} \cdot \frac{d}{dx} \sin(2x^3) \]

Step 2: Derive Sine Function

Next step is to derive the sine function. The derivative of \(\sin(x)\) is \(\cos(x)\) by trigonometric derivative rules. Applying this to \(\sin(2x^3)\), we get\[ \frac{d}{dx} \sin(2x^3) = \cos(2x^3) \cdot \frac{d}{dx} 2x^3 \]

Step 3: Derive Polynomial Function

Finally, we calculate the derivative of the polynomial function \(2x^3\). The rule states that the derivative of \(x^n\) is \(nx^{n-1}\). Applying this we get \[ \frac{d}{dx} 2x^3 = 6x^2 \]

Step 4: Combine all components

At the end, combine all the partial derivatives together as derived in steps 1, 2 and 3,\[ \frac{d}{dx} \sqrt{\sin(2x^3)} = \frac{1}{2 \sqrt{\sin(2x^3)}} \cdot \cos(2x^3) \cdot 6x^2 = 3x^2 \cdot \frac{\cos(2x^3)}{\sqrt{\sin(2x^3)}} \]

Key Concepts

Derivative of Composite Functions

Understanding the derivative of composite functions is essential when dealing with complex mathematical expresssions. In calculus, a composite function can be thought of as a function within another function. To differentiate such functions, we use the chain rule. The chain rule provides a method to break down the differentiation process into manageable parts by taking the derivative of the outer function and then multiplying it by the derivative of the inner function.

For example, consider the function \( f(g(x)) \), where \( g(x) \) is the inner function and \( f(u) \) (with \( u = g(x) \) ) is the outer function. The derivative of this composite function would be \( f'(g(x)) \) times \( g'(x) \). In the case of our exercise, \( \sqrt{\sin(2x^3)} \) is the composite function where \( \(\sin(2x^3)\) \) and \( \sqrt{u} \) represent the inner and outer functions, respectively. The chain rule simplifies the process and ensures a systematic approach to finding derivatives of even the most intricate functions.

Trigonometric Derivative Rules

When differentiating trigonometric functions, there are specific rules that one must follow. These trigonometric derivative rules help you find the rate of change of angles—in other words, how quickly or slowly the sine, cosine, or tangent of an angle changes as the angle itself changes. The derivative of \( \sin(x) \) is \( \cos(x) \) and the derivative of \( \cos(x) \) is \( -\sin(x) \). Similarly, other trigonometric functions like tangent and secant have their respective derivative rules.

In our example, we applied the rule for differentiating \( \sin(x) \) to the inner function \( 2x^3 \). The resulting derivative was \( \cos(2x^3) \) multiplied by the derivative of the polynomial function \( 2x^3 \), showing how the chain rule and trigonometric derivative rules work in tandem to solve more complex derivatives.

Derivative of Polynomial Functions

The derivative of polynomial functions follows a basic power rule that makes them relatively straightforward to differentiate compared to other types of functions. The power rule states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). In essence, you bring down the exponent as a coefficient and then subtract one from the original exponent.

In our exercise, we used this rule to differentiate \( 2x^3 \), producing a derivative of \( 6x^2 \) by bringing down the 3 as a coefficient (multiplying it by the existing 2) and subtracting one from the exponent. Polynomial functions are the building blocks of many mathematical expressions, and understanding how to differentiate them is vital in the broader scope of calculus.