Consider the function
y
(
x
)
=
tan
(
5
x
)
Find the derivative,
y
'
(
x
)
, and evaluate it at the
x
-coordinates in the table below. Fill out the table with your answers.

Answer
x
-coordinate	Derivative
x
1
=
−
0.11
y
'
(
x
1
)
=
x
2
=
0.05
y
'
(
x
2
)
=
x
3
=
0.13
y
'
(
x
3
)
=
Enter as many decimal places as your calculator allows (8 to 10). Your answer must be within
±
0.005
of the correct answer to be considered correct.

To find the derivative of the function \( y(x) = \tan(5x) \), we'll start by applying the chain rule. The chain rule states that if \( y = f(g(x)) \), then the derivative \( y' = f'(g(x)) \cdot g'(x) \).

### Step 1: Differentiate the Function

Given: 

\[
y(x) = \tan(5x)
\]

Using the chain rule, we differentiate:

\[
y'(x) = \frac{d}{dx} \tan(5x) = \sec^2(5x) \cdot \frac{d}{dx} (5x)
\]

Since the derivative of \( 5x \) is 5, this simplifies to:

\[
y'(x) = 5\sec^2(5x)
\]

### Step 2: Evaluate the Derivative at the Given Points

We need to evaluate \( y'(x) \) at the given \( x \)-coordinates:

1. \( x_1 = -0.11 \)
2. \( x_2 = 0.05 \)
3. \( x_3 = 0.13 \)

Let’s compute the values:

\[
y'(x_1) = 5\sec^2(5 \times (-0.11)) = 5\sec^2(-0.55)
\]

\[
y'(x_2) = 5\sec^2(5 \times 0.05) = 5\sec^2(0.25)
\]

\[
y'(x_3) = 5\sec^2(5 \times 0.13) = 5\sec^2(0.65)
\]

Let's calculate these values to as many decimal places as allowed:Here are the computed values for the derivatives at the given \(x\)-coordinates:

1. \( y'(x_1) = y'(-0.11) \approx 6.879490013 \)
2. \( y'(x_2) = y'(0.05) \approx 5.325997484 \)
3. \( y'(x_3) = y'(0.13) \approx 7.889553642 \)

These values are accurate to within 8 to 10 decimal places. 

Would you like more details on the calculations, or do you have any questions?

### Related Questions:
1. How do you apply the chain rule when differentiating other trigonometric functions?
2. What is the secant function, and why is it used in the derivative of a tangent function?
3. How does the value of the coefficient inside the tangent function affect its derivative?
4. Can you find the second derivative of the function \(y(x) = \tan(5x)\)?
5. How would you evaluate the derivative if the function were \(y(x) = \tan(kx)\) for any constant \(k\)?

**Tip:** Remember, the chain rule is essential when differentiating composite functions, especially those involving trigonometric expressions!

Learn how to find the derivative of y(x) = tan(5x) using the chain rule and evaluate it at specific x-coordinates: x1 = -0.11, x2 = 0.05, x3 = 0.13. This problem involves differentiating trigonometric functions and applying the chain rule in calculus.

Math Problem Statement

Consider the function y ( x )

Answer x -coordinate Derivative x 1

− 0.11 y ' ( x 1 )

x 2

0.05 y ' ( x 2 )

x 3

0.13 y ' ( x 3 )

Solution

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