Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Implicit Differentiation
Problem 3.8.50b
Textbook Question
45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
(x²+y²)²=25/4 xy²; (1, 2)

1
First, identify the given implicit equation of the curve: \((x^2 + y^2)^2 = \frac{25}{4} xy^2\). We need to find the derivative \(\frac{dy}{dx}\) using implicit differentiation.
Differentiate both sides of the equation with respect to \(x\). For the left side, use the chain rule: \(2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\). For the right side, apply the product rule: \(\frac{25}{4} (y^2 + 2xy \frac{dy}{dx})\).
Set the derivatives equal: \(2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = \frac{25}{4} (y^2 + 2xy \frac{dy}{dx})\).
Substitute the given point \((1, 2)\) into the differentiated equation to solve for \(\frac{dy}{dx}\). This will give the slope of the tangent line at the point.
Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step and \((x_1, y_1) = (1, 2)\), to write the equation of the tangent line.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. In this case, the equation (x² + y²)² = (25/4)xy² involves both x and y, requiring us to differentiate both sides with respect to x while treating y as a function of x. This method allows us to find dy/dx, which is essential for determining the slope of the tangent line.
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Tangent Line Equation
The equation of a tangent line at a given point on a curve can be expressed using the point-slope form: y - y₀ = m(x - x₀), where (x₀, y₀) is the point of tangency and m is the slope at that point. Once the derivative (slope) is calculated using implicit differentiation, it can be substituted into this formula along with the coordinates of the point (1, 2) to find the specific equation of the tangent line.
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Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. In the context of implicit differentiation, it allows us to differentiate terms involving y, which is a function of x. For example, when differentiating y², we apply the chain rule to obtain 2y(dy/dx), which is crucial for correctly finding the derivative of the given implicit equation.
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Intro to the Chain Rule
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