Problem 58 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( \frac{2x \cdot \csc(3-x^2) \cdot \left((1-x^2) \cdot \cot(3-x^2) + 1\right)}{(1-x^2)^2} \).

Step 1: Identify functions to differentiate

The function is given as \( f(x) = \frac{\csc(3-x^2)}{1-x^2} \). Recognize that it is a quotient of two functions: \( u(x) = \csc(3-x^2) \) in the numerator and \( v(x) = 1-x^2 \) in the denominator.

Step 2: Apply the Quotient Rule

To differentiate a quotient \( \frac{u}{v} \), use the Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \). Identify \( u(x) \), \( v(x) \), and find their derivatives \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate the numerator \( u(x) = \csc(3-x^2) \)

The derivative of \( \csc(g(x)) \) is \( -\csc(g(x)) \cdot \cot(g(x)) \cdot g'(x) \). Here, \( g(x) = 3-x^2 \) with derivative \( g'(x) = -2x \). Hence, \( u'(x) = -\csc(3-x^2) \cdot \cot(3-x^2) \cdot (-2x) = 2x \cdot \csc(3-x^2) \cdot \cot(3-x^2) \).

Step 4: Differentiate the denominator \( v(x) = 1-x^2 \)

The derivative of \( v(x) = 1-x^2 \) is \( v'(x) = -2x \).

Step 5: Substitute into the Quotient Rule

Substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into the Quotient Rule: \[\frac{d}{dx} \left( \frac{\csc(3-x^2)}{1-x^2} \right) = \frac{(1-x^2)(2x \cdot \csc(3-x^2) \cdot \cot(3-x^2)) - \csc(3-x^2)(-2x)}{(1-x^2)^2}\] This simplifies to:\[\frac{2x \cdot (1-x^2) \cdot \csc(3-x^2) \cdot \cot(3-x^2) + 2x \cdot \csc(3-x^2)}{(1-x^2)^2}\]

Step 6: Simplify the expression

Collect like terms: \[\frac{2x \cdot \csc(3-x^2) \cdot \left((1-x^2) \cdot \cot(3-x^2) + 1\right)}{(1-x^2)^2}\] This is the derivative of the function.

Key Concepts

Quotient Rule

When dealing with derivatives involving the division of two functions, the Quotient Rule is an essential tool. The rule is crucial for finding the derivative of a quotient like \( \frac{u}{v} \), where both \( u(x) \) and \( v(x) \) are differentiable functions. The Quotient Rule states:\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]Here's a step-by-step of how it works:

• Identify: Identify the two functions: the numerator \( u(x) \) and the denominator \( v(x) \).
• Differentiate: Differentiate both \( u(x) \) and \( v(x) \) to find \( u'(x) \) and \( v'(x) \).
• Substitute: Substitute these values into the Quotient Rule formula.
• Simplify: Simplify the resulting expression to get the derivative.

Knowing how to apply the Quotient Rule can make a tricky calculus problem much more manageable.

Trigonometric Functions

Trigonometric functions like sine, cosine, and cosecant often appear in calculus, especially in problems involving derivatives. In this exercise, we see the function \( \csc(x) \) being used. Here’s what you need to know:

• Basic Derivative Rules: The derivative of \( \csc(x) \) is crucial here. It is given by: \[ \frac{d}{dx}\left(\csc(x)\right) = -\csc(x) \cdot \cot(x) \]
• Combination with Quotient Rule: In our problem, \( \csc(3-x^2) \) forms part of a quotient, so we apply both the quotient rule and trigonometric derivative rules.

Understanding how trigonometric functions interact with derivative operations is essential in calculus, as it allows you to tackle varied types of problems effectively.

Chain Rule

The Chain Rule is a fundamental method in calculus used when taking derivatives of composite functions. A composite function is essentially a function within a function, like \( \csc(3-x^2) \) in our exercise. To successfully differentiate such functions, here's what you do:

• Identify the Outer and Inner Functions: Recognize the composite nature. Here, \( \csc \) is the outer function and \( 3-x^2 \) is the inner function \( g(x) \).
• Compute Derivatives: Find the derivative of the outer function as usual, then multiply by the derivative of the inner function \( g'(x) \). For \( g(x) = 3-x^2 \), we have \( g'(x) = -2x \).

The Chain Rule formula is:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]By mastering the Chain Rule, you can easily handle complex functions and break them down into more manageable parts.