write the eq of a line tangent to y=tan(2x) at x=pi/12

To find the equation of the line tangent to the function \( y = \tan(2x) \) at \( x = \frac{\pi}{12} \), follow these steps:

### 1. Find the derivative of \( y = \tan(2x) \)
The derivative of \( y = \tan(2x) \) gives the slope of the tangent line. Using the chain rule, we differentiate \( \tan(2x) \):

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

### 2. Evaluate the slope at \( x = \frac{\pi}{12} \)
Substitute \( x = \frac{\pi}{12} \) into the derivative to find the slope of the tangent line at this point:

\[
\frac{dy}{dx} \Bigg|_{x=\frac{\pi}{12}} = 2 \sec^2\left( 2 \times \frac{\pi}{12} \right) = 2 \sec^2\left(\frac{\pi}{6}\right)
\]

We know \( \sec\left(\frac{\pi}{6}\right) = 2 \), so:

\[
\frac{dy}{dx} \Bigg|_{x=\frac{\pi}{12}} = 2 \times 2^2 = 8
\]

The slope of the tangent line is 8.

### 3. Find the y-coordinate at \( x = \frac{\pi}{12} \)
Substitute \( x = \frac{\pi}{12} \) into the original function \( y = \tan(2x) \):

\[
y\left(\frac{\pi}{12}\right) = \tan\left(2 \times \frac{\pi}{12}\right) = \tan\left(\frac{\pi}{6}\right)
\]

We know \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \), so the y-coordinate at \( x = \frac{\pi}{12} \) is \( \frac{1}{\sqrt{3}} \).

### 4. Use the point-slope form to write the equation of the tangent line
The point-slope form of a line is:

\[
y - y_1 = m(x - x_1)
\]

where \( m \) is the slope and \( (x_1, y_1) \) is the point on the curve. We have \( m = 8 \), \( x_1 = \frac{\pi}{12} \), and \( y_1 = \frac{1}{\sqrt{3}} \). Substituting these values:

\[
y - \frac{1}{\sqrt{3}} = 8 \left( x - \frac{\pi}{12} \right)
\]

This is the equation of the tangent line.

Would you like more details or have any questions?

Here are some related questions you might want to explore:
1. How do you find the normal line to a curve?
2. What happens to the tangent line as the slope changes rapidly, like near asymptotes?
3. Can we extend this method to find tangent lines for other trigonometric functions?
4. How does the chain rule apply to differentiating composite functions in other contexts?
5. How does the derivative of a trigonometric function change with different amplitudes and frequencies?

**Tip:** When dealing with trigonometric functions, it's often helpful to review the values of sine, cosine, and tangent at key angles, such as \( \frac{\pi}{6}, \frac{\pi}{4}, \) and \( \frac{\pi}{3} \).

Learn how to find the equation of a tangent line to the function y = tan(2x) at x = π/12. This solution involves key calculus concepts like the derivative using the chain rule and applying the point-slope form for tangent lines. Suitable for students in Grades 11-12.

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