Math Problem Statement

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

Solution

The problem asks for the equation of the tangent line to the curve y=5x3+6y = \sqrt[3]{5x} + 6 at x=2x = 2.

Steps:

  1. Find the derivative: The slope of the tangent line is given by the derivative yy' of the function at x=2x = 2. The function is:

    y=5x3+6=(5x)1/3+6y = \sqrt[3]{5x} + 6 = (5x)^{1/3} + 6

    To differentiate this, we use the chain rule:

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

    Applying the chain rule:

    y=13(5x)2/35=53(5x)2/3y' = \frac{1}{3}(5x)^{-2/3} \cdot 5 = \frac{5}{3}(5x)^{-2/3}

  2. Evaluate the derivative at x=2x = 2: Now, substitute x=2x = 2 into the derivative:

    y(2)=53(52)2/3=53102/3y'(2) = \frac{5}{3}(5 \cdot 2)^{-2/3} = \frac{5}{3} \cdot 10^{-2/3}

    Simplify 102/310^{-2/3}:

    102/3=1102/310^{-2/3} = \frac{1}{10^{2/3}}

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

    m=53102/3m = \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=2x = 2, substitute x=2x = 2 into the original function:

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

    Therefore, the point is (2,103+6)(2, \sqrt[3]{10} + 6).

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

    yy1=m(xx1)y - y_1 = m(x - x_1)

    Substituting m=53102/3m = \frac{5}{3 \cdot 10^{2/3}}, x1=2x_1 = 2, and y1=103+6y_1 = \sqrt[3]{10} + 6:

    y(103+6)=53102/3(x2)y - (\sqrt[3]{10} + 6) = \frac{5}{3 \cdot 10^{2/3}}(x - 2)

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


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=5x4+6y = \sqrt[4]{5x} + 6?
  2. What is the general formula for the tangent line at any point x=ax = a on the curve y=5x3+6y = \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 x0x \to 0 for the function y=5x3+6y = \sqrt[3]{5x} + 6?

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

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Tangent Lines

Formulas

Derivative of (ax)^n: d/dx[(ax)^n] = n * a * (ax)^(n-1)
Point-slope form of a line: y - y1 = m(x - x1)

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12 (High School Calculus)