Problem 35 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\frac{\sec \left(x^{2}-1\right)}{\csc \left(x^{2}+1\right)} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{2x \cos(2)}{(\cos(x^2-1))^2} \).

Step 1: Rewrite the function

Start by rewriting the given function in terms of sine and cosine functions.\[f(x)=\frac{\sec(x^2-1)}{\csc(x^2+1)} = \frac{\frac{1}{\cos(x^2-1)}}{\frac{1}{\sin(x^2+1)}} = \frac{\sin(x^2+1)}{\cos(x^2-1)}\]

Step 2: Apply the Quotient Rule

Use the quotient rule for derivatives: if \(h(x) = \frac{u(x)}{v(x)}\), then \(h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\).Here, let \(u(x) = \sin(x^2+1)\) and \(v(x) = \cos(x^2-1)\).

Step 3: Differentiate u(x)

Find the derivative \(u'(x)\) using the chain rule. Let \(g(x) = x^2+1\), then \(u(x) = \sin(g(x))\).\[u'(x) = \cos(g(x)) \cdot g'(x) = \cos(x^2+1) \cdot 2x\]

Step 4: Differentiate v(x)

Find the derivative \(v'(x)\) using the chain rule. Let \(h(x) = x^2-1\), then \(v(x) = \cos(h(x))\).\[v'(x) = -\sin(h(x)) \cdot h'(x) = -\sin(x^2-1) \cdot 2x\]

Step 5: Substitute into the Quotient Rule

Substitute \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\) into the quotient rule formula.\[f'(x) = \frac{(\cos(x^2+1) \cdot 2x) \cdot \cos(x^2-1) - \sin(x^2+1) \cdot (-\sin(x^2-1) \cdot 2x)}{(\cos(x^2-1))^2}\]

Step 6: Simplify the Expression

Combine like terms and simplify the derivative expression.\[f'(x) = \frac{2x [\cos(x^2+1) \cdot \cos(x^2-1) + \sin(x^2+1) \cdot \sin(x^2-1)]}{(\cos(x^2-1))^2}\]Notice \(\cos(A)\cos(B) + \sin(A)\sin(B) = \cos(A-B)\) for angles, so:\[f'(x) = \frac{2x \cdot \cos((x^2+1)-(x^2-1))}{(\cos(x^2-1))^2} = \frac{2x \cdot \cos(2)}{(\cos(x^2-1))^2}\]

Key Concepts

Understanding Derivatives

In calculus, the concept of derivatives is crucial. They are used to determine the rate at which a quantity changes. Think of a derivative as the mathematical way of answering the question: "How fast is something happening?" For any given function, the derivative gives us the slope of the tangent line at any point. This tells us the function's rate of change at that point.

In our example, we have the function \[f(x)=\frac{\sec \left(x^{2}-1\right)}{\csc \left(x^{2}+1\right)}\] To find its derivative, we are calculating how this function changes as the variable \(x\) changes. By doing this, we can understand the behavior of the function more deeply, such as identifying if it's increasing or decreasing at different intervals.

Applying the Quotient Rule

The quotient rule is a handy tool when dealing with derivatives of functions that are the division of two separate functions. It helps us find derivatives in such composite structures.

Here's the basic idea: if you have two functions, \(u(x)\) and \(v(x)\), and you want to differentiate their quotient \(h(x) = \frac{u(x)}{v(x)}\), you use the formula:

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

In our example, \(u(x)\) is \(\sin(x^2+1)\) and \(v(x)\) is \(\cos(x^2-1)\). We find the derivatives of both \(u(x)\) and \(v(x)\), combine them using the quotient rule, and simplify. Applying the quotient rule allows us to break down the problem into manageable parts, making it easier to understand and solve.

Exploring Trigonometric Functions

Trigonometric functions are fascinating and widely used in calculus. They include sine, cosine, and tangent, each of which describes a relationship in a right-angle triangle. These functions are periodic, meaning they repeat their values in regular intervals, making them particularly useful for modeling cyclical phenomena.

In our exercise, trigonometric identities help simplify our expression. For instance, we can rewrite secant and cosecant using sine and cosine:

• \(\sec(\theta) = \frac{1}{\cos(\theta)}\)
• \(\csc(\theta) = \frac{1}{\sin(\theta)}\)

This transformation made it easier to find the derivatives using the quotient rule.

By understanding how trigonometric functions interplay, we can simplify complex problems and uncover elegant solutions, as illustrated in simplifying the inside of our derivative using the identity \(\cos(A)\cdot\cos(B) + \sin(A)\cdot\sin(B) = \cos(A-B)\). This powerful knowledge is invaluable in calculus and beyond.