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The problem asks for the equation of the tangent line to the curve \( y = \sqrt[3]{5x} + 6 \) at \( x = 2 \).

### Steps:

1. **Find the derivative**: 
   The slope of the tangent line is given by the derivative \( y' \) of the function at \( x = 2 \). The function is:

   \[
   y = \sqrt[3]{5x} + 6 = (5x)^{1/3} + 6
   \]

   To differentiate this, we use the chain rule:

   \[
   y' = \frac{d}{dx} \left( (5x)^{1/3} \right)
   \]

   Applying the chain rule:

   \[
   y' = \frac{1}{3}(5x)^{-2/3} \cdot 5 = \frac{5}{3}(5x)^{-2/3}
   \]

2. **Evaluate the derivative at \( x = 2 \)**: 
   Now, substitute \( x = 2 \) into the derivative:

   \[
   y'(2) = \frac{5}{3}(5 \cdot 2)^{-2/3} = \frac{5}{3} \cdot 10^{-2/3}
   \]

   Simplify \( 10^{-2/3} \):

   \[
   10^{-2/3} = \frac{1}{10^{2/3}}
   \]

   So, the slope of the tangent line at \( x = 2 \) is:

   \[
   m = \frac{5}{3 \cdot 10^{2/3}}
   \]

3. **Find the point on the curve**: 
   To find the point where the tangent touches the curve at \( x = 2 \), substitute \( x = 2 \) into the original function:

   \[
   y(2) = \sqrt[3]{5 \cdot 2} + 6 = \sqrt[3]{10} + 6
   \]

   Therefore, the point is \( (2, \sqrt[3]{10} + 6) \).

4. **Equation of the tangent line**:
   Using the point-slope form of the equation of a line:

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

   Substituting \( m = \frac{5}{3 \cdot 10^{2/3}} \), \( x_1 = 2 \), and \( y_1 = \sqrt[3]{10} + 6 \):

   \[
   y - (\sqrt[3]{10} + 6) = \frac{5}{3 \cdot 10^{2/3}}(x - 2)
   \]

This is the equation of the tangent line at \( x = 2 \).

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Would you like more details on any of these steps or explanations?

Here are 5 related questions for further practice:

1. How would the derivative change if the original function were \( y = \sqrt[4]{5x} + 6 \)?
2. What is the general formula for the tangent line at any point \( x = a \) on the curve \( y = \sqrt[3]{5x} + 6 \)?
3. What is the geometric interpretation of the derivative at a point on a curve?
4. How do we find the normal line (perpendicular) to a curve at a given point?
5. What happens to the slope of the tangent line as \( x \to 0 \) for the function \( y = \sqrt[3]{5x} + 6 \)?

**Tip**: Always check the units of each term in the derivative to avoid simplification errors when applying the chain rule!

Phương trình tiếp tuyến của đường cong y = ³√(5x) + 6 tại x = 2 là?

Learn how to find the equation of the tangent line to the curve y = ³√(5x) + 6 at x = 2. This problem involves differentiating using the chain rule and applying the point-slope form. A detailed, step-by-step solution is provided for high school students studying calculus.

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