Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 29-32, evaluate the indicated derivative. $$ F^{\prime}(1) \text { if } F(t)=\sin \left(t^{2}+3 t+1\right) $$

Step-by-Step Solution

Answer

The derivative \( F'(1) = 5\cos(5) \).

Step 1: Understand the problem

We are asked to find the derivative of the function \( F(t) = \sin(t^2 + 3t + 1) \) at \( t = 1 \). This means we need \( F'(t) \) and evaluate it at \( t = 1 \).

Step 2: Find the derivative using the chain rule

The function \( F(t) = \sin(u) \), where \( u = t^2 + 3t + 1 \). To find \( F'(t) \), use the chain rule: \( F'(t) = \cos(u) \cdot u' \). Find \( u' \) which is \( \frac{d}{dt}(t^2 + 3t + 1) = 2t + 3 \). Therefore, \( F'(t) = \cos(t^2 + 3t + 1) \cdot (2t + 3) \).

Step 3: Evaluate the derivative at t = 1

Now substitute \( t = 1 \) into the derived formula: \( F'(1) = \cos((1)^2 + 3(1) + 1) \cdot (2(1) + 3) \).

Step 4: Simplify the expression

Calculating the expression: \( (1)^2 + 3(1) + 1 = 5 \) and \( 2(1) + 3 = 5 \). So, \( F'(1) = \cos(5) \cdot 5 \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. A composite function has an "inner" function and an "outer" function. In this problem, the inner function is \( u = t^2 + 3t + 1 \) and the outer function is \( F(t) = \sin(u) \).
To apply the chain rule, you'll first differentiate the outer function with respect to the inner function \( u \), and then multiply that result by the derivative of the inner function \( u \) with respect to \( t \). This is expressed in the formula:

• Derivative of the outer function (sin): \( \frac{d}{du}(\sin(u)) = \cos(u) \).
• Derivative of the inner function (polynomial): \( \frac{d}{dt}(t^2 + 3t + 1) = 2t + 3 \).

Putting it all together, the derivative \( F'(t) \) is: \[ F'(t) = \cos(u) \cdot u' = \cos(t^2 + 3t + 1) \cdot (2t + 3) \].
The chain rule is crucial for working with more complex functions as it gives a systematic approach to differentiation.

Trigonometric Functions

Trigonometric functions such as sine (\( \sin \)) and cosine (\( \cos \)) appear frequently in calculus problems. These functions have specific rules when it comes to differentiation.
For the function \( F(t) = \sin(t^2 + 3t + 1) \), we focus on the derivative of \( \sin(u) \), which is \( \cos(u) \). This is consistent for any \( u \), where \( u \) is some function of \( t \).
Trigonometric derivatives, like those for sine and cosine, are straightforward:

• Derivative of \( \sin(u) \): \( \frac{d}{du}(\sin(u)) = \cos(u) \)
• Derivative of \( \cos(u) \): \( \frac{d}{du}(\cos(u)) = -\sin(u) \)

Remembering these basic rules helps in evaluating derivatives of functions that include trigonometric operations.

Evaluating Derivatives

Evaluating derivatives involves calculating the slope of the tangent line at a point on the function. In this exercise, after finding the general derivative using the chain rule, we plug in \( t = 1 \) to find \( F'(1) \).
This step asks us to substitute the value of \( t \) into our expression for \( F'(t) = \cos(t^2 + 3t + 1) \cdot (2t + 3) \):

• First, solve for \( (1)^2 + 3(1) + 1 = 5 \).
• Then, solve: \( 2(1) + 3 = 5 \).

Thus, \( F'(1) = \cos(5) \cdot 5 \).
This final result gives the rate of change of the function \( F(t) \) at \( t = 1 \), demonstrating how derivatives measure how a function changes as its input changes.