Math Problem Statement
Solution
The problem asks for the equation of the tangent line to the curve at .
Steps:
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Find the derivative: The slope of the tangent line is given by the derivative of the function at . The function is:
To differentiate this, we use the chain rule:
Applying the chain rule:
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Evaluate the derivative at : Now, substitute into the derivative:
Simplify :
So, the slope of the tangent line at is:
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Find the point on the curve: To find the point where the tangent touches the curve at , substitute into the original function:
[ y(2) = \sqrt[3]{5 \cdot 2} + 6 = \sqrt[3]{10} + 6 ]
Therefore, the point is .
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Equation of the tangent line: Using the point-slope form of the equation of a line:
Substituting , , and :
This is the equation of the tangent line at .
Would you like more details on any of these steps or explanations?
Here are 5 related questions for further practice:
- How would the derivative change if the original function were ?
- What is the general formula for the tangent line at any point on the curve ?
- What is the geometric interpretation of the derivative at a point on a curve?
- How do we find the normal line (perpendicular) to a curve at a given point?
- What happens to the slope of the tangent line as for the function ?
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)