Problem 9 - Calculus Single Variable

Question

Compute the indicated derivative for the given function by using the formulas and rules that are summarized at the end of this section. $$ \left.\frac{d g}{d s}\right|_{s=-2}, g(s)=3 s^{3}-\cos (s) $$

Step-by-Step Solution

Answer

The derivative \( \left.\frac{d g}{d s}\right|_{s=-2} = 35.091 \).

Step 1: Find the Derivative

First, find the derivative of the function \( g(s) = 3s^3 - \cos(s) \). Use the power rule and the derivative of cosine:- The power rule states \( \frac{d}{ds}(s^n) = n \cdot s^{n-1} \).- The derivative of \( \cos(s) \) is \( -\sin(s) \).Thus, the derivative \( g'(s) = \frac{d}{ds}(3s^3) - \frac{d}{ds}(\cos(s)) = 3 \cdot 3s^2 + \sin(s) = 9s^2 + \sin(s) \).

Step 2: Evaluate the Derivative

Now that you have the derivative, evaluate it at \( s = -2 \). Substitute \( -2 \) into the derivative:\[ g'(-2) = 9(-2)^2 + \sin(-2) \].

Step 3: Compute Values

Compute the values from the expression:- \((-2)^2 = 4\), so \(9 \times 4 = 36\).- Use a calculator to find \( \sin(-2) \approx -0.909\).Thus, \[ g'(-2) = 36 - 0.909 = 35.091 \].

Key Concepts

Power Rule

When calculating derivatives, one of the most powerful and frequently used tools is the power rule. This rule provides a quick way to differentiate functions where the variable is raised to a power. According to the power rule, the derivative of a function in the form of \( s^n \) is given by \( \frac{d}{ds}(s^n) = n \cdot s^{n-1} \). In simpler terms, you bring the exponent down in front of the variable and then reduce the exponent by one.

For example, in the function \( g(s) = 3s^3 \), you would first bring the 3 down to multiply by the coefficient (which is also 3 here), making it \( 3 \cdot 3 \). Then, you reduce the power from 3 to 2, giving you \( 9s^2 \). This transformation simplifies the process of differentiation markedly, especially with polynomial functions.

Trigonometric Derivatives

Trigonometric functions such as sine and cosine have specific derivatives that are essential to remember when dealing with calculus. The derivatives of trigonometric functions often appear in problems that involve curves or periodic phenomena.

In the context of this exercise, you work with the function \(-\cos(s)\). The derivative of \( \cos(s) \) is \(-\sin(s)\). This is a crucial standard result: remember that when you take the derivative of the cosine function, you swap cosine for sine and introduce a negative sign. Thus, when applying this derivative, the cosine becomes the sine, but you must include that negative before the sine, resulting in \(-\sin(s)\).

Understanding these foundational trigonometric derivatives helps when combining them in functions, just like in this exercise where they appear alongside polynomial derivatives. This versatility makes these derivatives valuable tools in your calculus toolkit.

Evaluation of Derivatives

After finding the derivative of a function, the next logical step is often to evaluate it at a specific point. This involves substituting a particular value of the variable into the derivative to find the slope of the tangent at that point.

For this exercise, once the derivative \( g'(s) = 9s^2 + \sin(s) \) is found, the task is to evaluate it at \( s = -2 \). This means substituting \(-2\) into the derivative function: \( g'(-2) = 9(-2)^2 + \sin(-2) \). Here, \((-2)^2 = 4\), which results in \( 9 \times 4 = 36 \). Then, using a calculator or reference, you find \( \sin(-2) \approx -0.909 \).

• Substitute \(-2\) into the polynomial part: \( 9 \times (-2)^2 = 36 \).
• Find \( \sin(-2) \) which is approximately \(-0.909\).
• Add the values: \( 36 + (-0.909) = 35.091 \).

This method of evaluation offers insight into the behavior of the original function at specific values, giving practical meaning to its rate of change.