Problem 30 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ g(s)=\left(\cos ^{2} s-3 s^{2}\right)^{2} $$

Step-by-Step Solution

Answer

The derivative is \( g'(s) = 2(\cos^2 s - 3s^2)(-\sin(2s) - 6s) \).

Step 1: Understand the Function

The function is given as \( g(s) = \left(\cos^2 s - 3s^2\right)^2 \). It is a composition of functions, specifically a power function applied to another function, which itself is a combination of trigonometric and polynomial expressions.

Step 2: Apply the Chain Rule

To find the derivative of \( g(s) = \left( u(s) \right)^2 \), where \( u(s) = \cos^2 s - 3s^2 \), we use the chain rule. The chain rule states that \( g'(s) = 2u(s) \cdot u'(s) \). So, we first need \( u'(s) \).

Step 3: Differentiate the Inner Function

Now differentiate the inner function \( u(s) = \cos^2 s - 3s^2 \). Using product and chain rules, the derivative of \( \cos^2 s \) is \( -2\cos s\sin s = -\sin (2s) \), and for \( -3s^2 \) is \( -6s \). Therefore, \( u'(s) = -\sin(2s) - 6s \).

Step 4: Combine Results Using the Chain Rule

Substitute \( u(s) = \cos^2 s - 3s^2 \) and \( u'(s) = -\sin(2s) - 6s \) into the chain rule result: \[ g'(s) = 2(\cos^2 s - 3s^2)(-\sin(2s) - 6s) \].

Step 5: Simplify the Expression

Distribute the term \( 2(\cos^2 s - 3s^2) \) within the parentheses: \[ g'(s) = 2(\cos^2 s - 3s^2)(-\sin(2s)) + 2(\cos^2 s - 3s^2)(-6s) \]. Further simplify if needed, focusing on coefficient distribution and any further factoring or consolidation.

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of a composition of functions. When you have a function nestled inside another, the chain rule helps us differentiate it effectively. Here's how it works in simple terms: If you have a composite function like our example, \( g(s) = (u(s))^2 \), where \( u(s) = \cos^2 s - 3s^2 \), the chain rule lets you differentiate the outer function and multiply it by the derivative of the inside function.In this case:- First, identify the outside and inside functions.- The outside function is a power function, specifically squared, resulting in \( 2\cdot (u(s)) \). - Multiply by the derivative of the inside function \( u'(s) \).Applying the chain rule, we get:\[g'(s) = 2(u(s))\cdot u'(s) \]It may initially seem complex, but with practice, the chain rule becomes an invaluable part of finding derivatives quickly.

Trigonometric Differentiation

Trigonometric differentiation involves finding the derivative of trigonometric functions. In calculus, these derivatives require unique rules that relate to the unit circle and periodicity of trig functions.Consider the trigonometric function in our example:- We have the function \( \cos^2 s \). This can be thought of as \( (\cos s)^2 \), which makes it a composite function itself.- Using the chain rule again, the derivative of \( \cos s \) and squaring it gives us the formula: \[\frac{d}{ds} (\cos^2 s) = -2\cdot \cos s\cdot \sin s = -\sin(2s)\]Trigonometric derivatives often require leveraging identities, such as the double angle identity, to simplify results.- Remember common derivatives:

• The derivative of \( \sin(x) \) is \( \cos(x) \)
• The derivative of \( \cos(x) \) is \(-\sin(x) \)

Trigonometric differentiation is crucial for solving problems involving oscillatory behavior, waves, and many practical applications in physics and engineering.

Polynomial Functions

Polynomial functions are some of the simplest and most common functions you'll encounter, defined by expressions like \( -3s^2 \). Differentiating polynomial functions involves basic rules that are easy to apply:- The power rule states that for any function \( ax^n \), the derivative is \( n\cdot ax^{n-1} \).Let's apply this to part of our function:- Given \( -3s^2 \), we use the power rule: \[\frac{d}{ds} (-3s^2) = -6s\]Polynomials are straightforward because the rules are consistent:

• Derivatives of constant terms are zero.
• Simplifying derivatives involves combining like terms.

Understanding how to handle polynomial functions is a foundation in calculus that leads into more complex topics seamlessly. This knowledge provides strong groundwork for tackling more involved mathematical problems.