Problem 47 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ g(x)=\frac{1}{\csc ^{3}\left(1-5 x^{2}\right)} $$

Step-by-Step Solution

Answer

The derivative is \( 30x \csc^5(1-5x^2) \cot(1-5x^2) \).

Step 1: Understand the Function

The given function is \( g(x) = \frac{1}{\csc^3(1-5x^2)} \). This can be rewritten as \( g(x) = \csc^{-3}(1-5x^2) \).

Step 2: Apply the Chain Rule

To differentiate \( g(x) = [\csc(1-5x^2)]^{-3} \), recognize it as a composition of functions. Begin by using the chain rule: if \( y = [u(x)]^n \), then \( \frac{dy}{dx} = n[u(x)]^{n-1} \cdot u'(x) \). Here, \( u(x) = \csc(1-5x^2) \) and \( n = -3 \).

Step 3: Differentiate the Outer Function

Differentiate the outer function with respect to \( u \): \( \frac{d}{du}[u]^{-3} = -3u^{-4} \cdot u' \). Thus, \( \frac{d}{dx}[\csc(1-5x^2)]^{-3} = -3[\csc(1-5x^2)]^{-4} \cdot \frac{d}{dx}[\csc(1-5x^2)] \).

Step 4: Differentiate the Inner Function

Use the chain rule again for \( u(x) = \csc(1-5x^2) \). The derivative of \( \csc(v) \) with respect to \( v \) is \( -\csc(v)\cot(v) \). Let \( v = 1-5x^2 \), so \( \frac{dv}{dx} = -10x \). Hence, \( \frac{d}{dx}[\csc(1-5x^2)] = -\csc(1-5x^2)\cot(1-5x^2) \cdot (-10x) \).

Step 5: Combine Derivatives

Substitute \( \frac{d}{dx}[\csc(1-5x^2)] \) back into the derivative from Step 3: \[\frac{d}{dx}[\csc(1-5x^2)]^{-3} = -3[\csc(1-5x^2)]^{-4} \cdot (-10x) \cdot \csc(1-5x^2)\cot(1-5x^2)\]Simplify to get: \[ g'(x) = 30x \csc^5(1-5x^2) \cot(1-5x^2)\]

Step 6: Conclusion: Write the Final Derivative

The derivative of the given function is \[ g'(x) = 30x \csc^5(1-5x^2) \cot(1-5x^2)\]

Key Concepts

Derivatives in Calculus

Derivatives play a crucial role in calculus and allow us to understand how functions change. At its core, the derivative measures the rate at which a function's value changes as its input changes. This concept is fundamental in many fields, from physics to economics.

• To calculate a derivative, we explore the limit of how much a function changes over an infinitesimally small interval.
• Mathematically, for a function \( f(x) \), its derivative \( f'(x) \) is given by \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).

The derivative tells us the slope of the tangent line to the curve at any given point.
In situations where functions are represented as a composition of other functions, the chain rule becomes especially useful.
Understanding derivatives deeply helps in reflecting on changes, trends, and behaviors of different systems or models displayed by the function.

Understanding the Chain Rule

The chain rule is an essential technique in calculus for finding the derivative of composite functions. A composite function combines two or more functions, which makes direct differentiation challenging. The chain rule addresses this problem by breaking the process into manageable parts.

• If you have a function \( y = f(g(x)) \), the chain rule tells us to first take the derivative of \( f(u) \) with respect to \( u \) (where \( u = g(x) \)) and multiply it by the derivative of \( g(x) \).
• Mathematically, this is expressed as \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \).

The chain rule is particularly beneficial when dealing with functions nested within trigonometric, exponential, or logarithmic functions.
This strategy was crucial in solving the original exercise, where the trigonometric function cosecant was expressed with a power minus three applied, requiring successive differentiation steps.
Implementing the chain rule helps to systematically differentiate complex combinations efficiently, leading to cleaner and more comprehensible solutions.

Exploring Trigonometric Functions

Trigonometric functions like sine, cosine, and cosecant are foundational in calculus, as they describe relationships between the angles and sides of triangles. These functions not only apply to geometry but also model periodic phenomena like sound and light waves.

• In calculus, trigonometric functions have distinct derivatives:
• For example, if \( f(x) = \sin(x) \), then \( f'(x) = \cos(x) \).
• If \( f(x) = \cos(x) \), then \( f'(x) = -\sin(x) \).
• For \( f(x) = \csc(x) \), the derivative is \( f'(x) = -\csc(x)\cot(x) \).

Understanding how to differentiate trigonometric functions helps to solve more complex problems involving them.
In the given exercise, differentiating the cosecant function formed an integral part of applying the chain rule.
Quickly recognizing these derivative patterns enables more efficient problem-solving and prepares students for advanced topics in calculus.