Math Problem Statement
Solution
The problem asks for the equation of the tangent line to the curve at the point , using implicit differentiation.
We need to implicitly differentiate both sides of the given equation with respect to and then solve for the slope at the point .
Step 1: Differentiate both sides of the equation
We start with: Differentiate each term with respect to , using the product rule where necessary:
- For , apply the product rule: This simplifies to .
- .
So the differentiated equation is:
Step 2: Substitute the point
Substitute and into the equation to find :
- For , we get ,
- For , we get ,
- For , we get ,
- For , we get .
Substituting into the differentiated equation gives: Simplify: Solve for :
Step 3: Equation of the tangent line
The slope of the tangent line is , and the point on the curve is . Using the point-slope form of the equation of a line:
Simplify:
Thus, the equation of the tangent line is:
Let me know if you'd like to go over any steps or explore related concepts!
Here are some follow-up questions:
- What is implicit differentiation, and why is it used here?
- How is the product rule applied when differentiating terms like ?
- Why do we substitute the point after differentiating?
- How do we simplify expressions involving fractions, like in the final equation of the tangent line?
- What are other methods to find the equation of a tangent line, and when do they apply?
Tip: When differentiating implicitly, remember to apply the chain rule whenever you differentiate terms involving , since is a function of .
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Math Problem Analysis
Mathematical Concepts
Implicit Differentiation
Tangent Line
Calculus
Formulas
Implicit Differentiation
Product Rule: d(uv)/dx = u'(v) + u(v')
Equation of a Line: y - y1 = m(x - x1)
Theorems
Chain Rule: d(f(g(x)))/dx = f'(g(x))g'(x)
Implicit Differentiation Theorem
Suitable Grade Level
Grades 11-12 or Introductory College Calculus
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