Problem 46 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( g'(x) = 10 \cot(5x) \csc^2(5x) \).

Step 1: Review the Function

We need to find the derivative of the function \( g(x) = \frac{1}{\csc^2(5x)} \). This can be rewritten as \( g(x) = \csc^{-2}(5x) \).

Step 2: Apply the Chain Rule

To differentiate \( g(x) = \csc^{-2}(5x) \), first note that it is composed of the outer function \( f(u) = u^{-2} \) where \( u = \csc(5x) \). The derivative of \( f(u) = u^{-2} \) is \( f'(u) = -2u^{-3} \).

Step 3: Differentiate the Inner Function

Calculate the derivative of the inner function \( \csc(5x) \). The derivative of \( \csc(x) \) is \( -\csc(x)\cot(x) \), so for \( \csc(5x) \), use the chain rule to get \( -5\csc(5x)\cot(5x) \).

Step 4: Combine the Derivatives Using the Chain Rule

Using the chain rule, combine the derivatives: \[ g'(x) = f'(u) \cdot u'(x) = (-2)(\csc(5x))^{-3} \cdot (-5\csc(5x)\cot(5x)) \].

Step 5: Simplify the Expression

Simplify the derivative: \[ g'(x) = 10\csc^{-3}(5x) \cdot \csc(5x) \cdot \cot(5x) = 10\csc^{-2}(5x) \cdot \cot(5x) \].

Step 6: Final Result

The derivative of \( g(x) = \frac{1}{\csc^2(5x)} \) is: \[ g'(x) = 10 \cot(5x) \csc^2(5x) \].

Key Concepts

Understanding the Chain Rule

One of the fundamental techniques for finding derivatives in calculus is the Chain Rule. This rule is especially useful when dealing with complex functions that are compositions of two or more simpler functions. Essentially, the Chain Rule allows us to take the derivative of a composition by breaking it down into an *outer function* and an *inner function*.
To apply the Chain Rule:

• Identify the outer function, which is the function applied directly to the expression.
• Identify the inner function, which is inside the outer function.
• Differentiate each part separately. First, take the derivative of the outer function with respect to the inner function. Then, differentiate the inner function with respect to the original variable.
• Multiply these derivatives together to find the final derivative.

This method simplifies the differentiation of functions like the one in our example: \[ g(x)= \csc^{-2}(5x) \] where the outer function is \( u^{-2} \) and the inner function is \( u = \csc(5x) \). Applying the Chain Rule can make the process much more manageable.

Exploring Trigonometric Functions

Trigonometric functions like \( \sin, \cos, \tan \), and \( \csc \) are foundational in calculus and have unique derivatives. Understanding these will greatly aid in solving derivative problems involving trigonometric identities. For instance, the derivative of the cosecant function \( \csc(x) \) is \( -\csc(x)\cot(x) \). This means it incorporates other trigonometric functions within its derivative.
In our problem, the derivative \( \csc(5x) \) requires using this specific derivative rule because the argument is more complex than a simple \( x \). In this case, the Chain Rule helps us adjust for the compound argument \( 5x \). By applying the appropriate derivative formula and then adjusting it through multiplication by the derivative of the inner expression \( 5 \), we end up with:\[ -5 \csc(5x) \cot(5x) \].

Simplifying Derivatives

Simplification is often the final step in finding derivatives, turning a complex expression into something more usable. After applying the Chain Rule and calculating the derivatives of both the outer and inner functions, you're left with a product that can often be simplified by combining or canceling terms.
In simplifying the expression \[ g'(x) = 10 \csc^{-3}(5x) \cdot \csc(5x) \cdot \cot(5x) \],we combine the cosecants:

• Multiply \( \csc^{-3}(5x) \) by \( \csc(5x) \) to get \( \csc^{-2}(5x) \).

This reduces our expression to:\[ g'(x) = 10 \cot(5x) \csc^2(5x) \],providing a streamlined form of the derivative. Placing the derivative in its simplest form helps make the subsequent steps or evaluations quicker and more accurate.