Problem 39 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 6x [\sec^2(3x^2 - 1) \cot(3x^2 + 1) - \tan(3x^2 - 1) \csc^2(3x^2 + 1)] \).

Step 1: Identify Function Types

We are asked to find the derivative of the function \( f(x) = \tan(3x^2 - 1) \cdot \cot(3x^2 + 1) \). Notice that this is a product of two trigonometric functions. To differentiate this, we will need to use both the product rule and the chain rule.

Step 2: Apply the Product Rule

The product rule states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product is \( u'(x)v(x) + u(x)v'(x) \). Set \( u(x) = \tan(3x^2 - 1) \) and \( v(x) = \cot(3x^2 + 1) \). First, find \( u'(x) \) and \( v'(x) \) using the chain rule.

Step 3: Differentiate \( u(x) = \tan(3x^2 - 1) \) Using Chain Rule

To differentiate \( u(x) = \tan(3x^2 - 1) \), apply the chain rule. The derivative of \( \tan(x) \) is \( \sec^2(x) \). Therefore, \( u'(x) = \sec^2(3x^2 - 1) \cdot \frac{d}{dx}(3x^2 - 1) = \sec^2(3x^2 - 1) \cdot 6x \).

Step 4: Differentiate \( v(x) = \cot(3x^2 + 1) \) Using Chain Rule

To differentiate \( v(x) = \cot(3x^2 + 1) \), apply the chain rule. The derivative of \( \cot(x) \) is \( -\csc^2(x) \). Therefore, \( v'(x) = -\csc^2(3x^2 + 1) \cdot \frac{d}{dx}(3x^2 + 1) = -\csc^2(3x^2 + 1) \cdot 6x \).

Step 5: Substitute into Product Rule

Using the derivatives found: \( u'(x) = 6x \sec^2(3x^2 - 1) \) and \( v'(x) = -6x \csc^2(3x^2 + 1) \), substitute back into the product rule formula: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \].

Step 6: Calculate \( f'(x) \)

Substitute the expressions: \[ f'(x) = (6x \sec^2(3x^2 - 1)) \cdot \cot(3x^2 + 1) + \tan(3x^2 - 1) \cdot (-6x \csc^2(3x^2 + 1)) \]. Simplify the expression to get the final derivative.

Key Concepts

Product Rule

When we differentiate the product of two functions, such as \(f(x) = \tan(3x^2 - 1) \cdot \cot(3x^2 + 1)\), we apply the product rule. The product rule is a fundamental tool in calculus for finding derivatives of product expressions.

It states that if you have two functions, \(u(x)\) and \(v(x)\), the derivative of their product is given by:

• \(u'(x) v(x) + u(x) v'(x)\)

This means that you need to take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

In our problem:

• Let \(u(x) = \tan(3x^2 - 1)\)
• and \(v(x) = \cot(3x^2 + 1)\).

First, we find \(u'(x)\) and \(v'(x)\) using other rules of differentiation, such as the chain rule, which is explained next.

Chain Rule

The chain rule is essential when differentiating composite functions, such as \(\tan(3x^2 - 1)\) and \(\cot(3x^2 + 1)\).

This rule allows us to find the derivative of compositions of functions by differentiating the outer function and then multiplying by the derivative of the inner function. Mathematically, if you have a composite function \(g(f(x))\), the chain rule is expressed as:

• \( g'(f(x)) \cdot f'(x) \)

For example, in differentiating \(\tan(3x^2 - 1)\), we:

• Differentiate the outer function: \(\tan(x)\) becomes \(\sec^2(x)\).
• Multiply by the derivative of the inner function: \((3x^2 - 1)' = 6x\).

So, \(u'(x) = \sec^2(3x^2 - 1) \cdot 6x\).

Similarly, \(v'(x)\) for \(\cot(3x^2 + 1)\) involves differentiating the outer function \(\cot(x)\) to get \(-\csc^2(x)\), and multiplying by the derivative of the inner function \((3x^2 + 1)' = 6x\). Thus, \(v'(x) = -\csc^2(3x^2 + 1) \cdot 6x\).

Trigonometric Functions

Trigonometric functions such as \(\tan(x)\) and \(\cot(x)\) play a significant role in calculus, especially in problems involving rates of change and periodic patterns.

Understanding their derivatives is crucial.

The derivative of \(\tan(x)\) is \(\sec^2(x)\). This comes from the identity that relates tangent and secant functions. Likewise, \(\cot(x)\) has a derivative of \(-\csc^2(x)\), which involves the cosecant function that is the reciprocal of sine.

These relationships highlight how trigonometric identities and derivatives are intertwined. They also emphasize the importance of memorizing certain derivative formulas for efficient problem-solving in calculus.

In our example, these trigonometric derivatives are applied within the context of the chain rule to find \(u'(x)\) and \(v'(x)\), which are essential for applying the product rule and calculating the final result.