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Solve this: Consider the function y(x)=tan(2x)y(x)=tan(2x)

Find the derivative, y'(x)y′(x), and evaluate it at the xx-coordinates in the table below. Fill out the table with your answers.

Answer

xx-coordinateDerivative

x1=−0.41x1=-0.41

y'(x1)y′(x1) =

x2=−0.04x2=-0.04

y'(x2)y′(x2) =

x3=0.31x3=0.31

y'(x3)y′(x3) =

Enter *as many decimal places as your calculator allows* (8 to 10). Your answer must be within ±0.005±0.005 of the correct answer to be considered correct.

To solve the problem, we need to find the derivative of the function \( y(x) = \tan(2x) \) and evaluate it at the specified \( x \)-coordinates.

### Step 1: Find the Derivative \( y'(x) \)

The function given is:

\[
y(x) = \tan(2x)
\]

To find the derivative \( y'(x) \), we apply the chain rule. The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \), and the derivative of \( 2x \) with respect to \( x \) is 2. Thus:

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

### Step 2: Evaluate the Derivative at the Given \( x \)-coordinates

Now, let's evaluate \( y'(x) = 2 \sec^2(2x) \) at the points \( x_1 = -0.41 \), \( x_2 = -0.04 \), and \( x_3 = 0.31 \).

1. **For \( x_1 = -0.41 \):**

\[
y'(-0.41) = 2 \sec^2(2 \cdot -0.41) = 2 \sec^2(-0.82)
\]

\[
\sec(-0.82) = \frac{1}{\cos(-0.82)}
\]

Calculating the cosine:

\[
\cos(-0.82) \approx 0.68192
\]

Then:

\[
\sec^2(-0.82) = \left( \frac{1}{0.68192} \right)^2 \approx 2.148
\]

Thus:

\[
y'(-0.41) = 2 \times 2.148 \approx 4.296
\]

2. **For \( x_2 = -0.04 \):**

\[
y'(-0.04) = 2 \sec^2(2 \cdot -0.04) = 2 \sec^2(-0.08)
\]

\[
\cos(-0.08) \approx 0.9968
\]

\[
\sec^2(-0.08) = \left( \frac{1}{0.9968} \right)^2 \approx 1.00643
\]

Thus:

\[
y'(-0.04) = 2 \times 1.00643 \approx 2.01286
\]

3. **For \( x_3 = 0.31 \):**

\[
y'(0.31) = 2 \sec^2(2 \cdot 0.31) = 2 \sec^2(0.62)
\]

\[
\cos(0.62) \approx 0.81342
\]

\[
\sec^2(0.62) = \left( \frac{1}{0.81342} \right)^2 \approx 1.5102
\]

Thus:

\[
y'(0.31) = 2 \times 1.5102 \approx 3.0204
\]

### Step 3: Final Results

- \( y'(-0.41) \approx 4.296 \)
- \( y'(-0.04) \approx 2.01286 \)
- \( y'(0.31) \approx 3.0204 \)

These are the derivative values at the specified \( x \)-coordinates.

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

### Related Questions

1. How would the derivative change if the function was \( y(x) = \tan(3x) \)?
2. What is the second derivative of \( y(x) = \tan(2x) \)?
3. How do you compute the secant function for any angle?
4. What happens to the value of the derivative as \( x \) approaches \( \frac{\pi}{4} \)?
5. What is the behavior of the function \( \tan(2x) \) and its derivative as \( x \) approaches infinity?

**Tip:** Remember that the secant function \( \sec(x) \) is undefined where \( \cos(x) = 0 \). Be careful when evaluating trigonometric functions at critical points!

Learn how to calculate the derivative of y(x) = tan(2x) using the chain rule in trigonometry. Evaluate the derivative at x = -0.41, x = -0.04, and x = 0.31 with step-by-step solutions and accurate decimal approximations.

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