Problem 58 - Calculus for Biology and Medicine

Question

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 of \( f(x) = \frac{\csc(3-x^2)}{1-x^2} \) is \( f'(x) = \frac{2x \csc(3-x^2) [\cot(3-x^2) (1-x^2) + 1]}{(1-x^2)^2} \).

Step 1: Identify Function Components

We are given the function \( f(x) = \frac{\csc(3-x^2)}{1-x^2} \). This is a quotient of two functions: the numerator \( u(x) = \csc(3-x^2) \) and the denominator \( v(x) = 1-x^2 \). We will need to apply the quotient rule for derivatives since these are two functions divided by one another.

Step 2: Recall the Quotient Rule

The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then the derivative is given by \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \). We will apply this rule to the given function.

Step 3: Differentiate the Numerator

First, find the derivative of \( u(x) = \csc(3-x^2) \). Recall that the derivative of \( \csc(x) \) is \(-\csc(x)\cot(x)\). Using the chain rule, the derivative is \( u'(x) = -\csc(3-x^2) \cot(3-x^2) \times (-2x) = 2x \csc(3-x^2)\cot(3-x^2) \).

Step 4: Differentiate the Denominator

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

Step 5: Apply the Quotient Rule

Substitute \( u'(x) = 2x \csc(3-x^2) \cot(3-x^2) \), \( u(x) = \csc(3-x^2) \), \( v'(x) = -2x \), and \( v(x) = 1-x^2 \) into the quotient rule formula. This gives: \[ f'(x) = \frac{(2x \csc(3-x^2) \cot(3-x^2))(1-x^2) - (\csc(3-x^2))(-2x)}{(1-x^2)^2}\]

Step 6: Simplify the Expression

Simplify the numerator of the expression: \[ f'(x) = \frac{2x \csc(3-x^2) \cot(3-x^2) (1-x^2) + 2x \csc(3-x^2)}{(1-x^2)^2}\]Factor out the common term \(2x \csc(3-x^2)\):\[f'(x) = \frac{2x \csc(3-x^2) [\cot(3-x^2) (1-x^2) + 1]}{(1-x^2)^2}\]

Key Concepts

Understanding the Quotient Rule

When you encounter a function that is the division of two expressions, like our problem here, the ideal choice is to apply the quotient rule. The rule simplifies finding derivatives of functions expressed as quotients. Formally expressed, if we have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable functions, the derivative is:

• \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)

It's crucial first to identify the components \( u(x) \), the numerator, and \( v(x) \), the denominator. Then you differentiate each part, respectively, \( u'(x) \) for the numerator and \( v'(x) \) for the denominator. Finally, substitute these derivatives into the quotient rule formula and simplify. This method is especially helpful when dealing with complex expressions in calculus.

Applying the Chain Rule

In calculus, the chain rule is essential when you have a composite function — a function within another function. For instance, in the function \( u(x) = \csc(3-x^2) \), the argument \( 3-x^2 \) is a function itself that multiplies with the outer function \( \csc \).The chain rule asserts that the derivative of a composite function \( f(g(x)) \) is the derivative of the outer function, \( f'(g(x)) \), multiplied by the derivative of the inner function, \( g'(x) \). Hence for our example, the derivative is calculated as:

• \( u'(x) = -\csc(3-x^2)\cot(3-x^2) \times (-2x) \)

This chaining of derivatives makes it possible to tackle more complex derivatives where basic rules wouldn't suffice, streamlining the solution process significantly.

Exploring Trigonometric Differentiation

Trigonometric functions, like sine, cosine, and cosecant, often appear in calculus problems. Their differentiation requires certain memorized derivatives:

• \( \sin(x) \to \cos(x) \)
• \( \cos(x) \to -\sin(x) \)
• \( \csc(x) \to -\csc(x)\cot(x) \)

Trigonometric differentiation plays a pivotal role in understanding how rates change across angles and cycles. In our example, differentiating \( \csc(3-x^2) \) required combining the basic trigonometric differentiation rule with the chain rule to account for the internal function \( 3-x^2 \). Knowing these rules aids in efficiently solving trigonometric-centered calculus problems.

The Art of Calculus Problem Solving

Approaching calculus problems involves a strategic mix of rules, logical steps, and simplification. Here's a brief on how a problem like this unfolds: 1. **Identify the Structure**: Recognize the function's structure — is it a quotient, composite, or trigonometric identity? 2. **Choose the Right Tool**: Use the quotient rule for division, the chain rule for composites, and trigonometric rules as needed. 3. **Calculate Derivatives**: Differentiate each component involved. 4. **Combine and Simplify**: Merge the differentiated parts back into the quotient formula. 5. **Simplify Again**: Clean up any algebraic fractions or terms to obtain the simplest form. By following these steps, even seemingly complex calculus problems become manageable. It boils down to understanding which rules apply and how they interconnect, ensuring clarity and accuracy in deriving solutions.