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.R.56
Textbook Question
Evaluate and simplify y'.
xy⁴+x⁴y=1

1
First, recognize that the given equation is implicit: \( y \cdot x y^4 + x^4 y = 1 \). We need to differentiate both sides with respect to \( x \).
Apply the product rule to the term \( y \cdot x y^4 \). The product rule states that \( (uv)' = u'v + uv' \). Here, let \( u = y \) and \( v = x y^4 \). Differentiate \( u \) and \( v \) with respect to \( x \).
Differentiate \( u = y \) with respect to \( x \) to get \( u' = y' \). For \( v = x y^4 \), apply the product rule again: \( v' = (x)' y^4 + x (y^4)' \). Differentiate \( x \) to get 1, and \( y^4 \) using the chain rule to get \( 4y^3 y' \).
Now, differentiate the second term \( x^4 y \) using the product rule: \( (x^4 y)' = (x^4)' y + x^4 (y)' \). Differentiate \( x^4 \) to get \( 4x^3 \), and \( y \) to get \( y' \).
Combine all differentiated terms and set the derivative of the right side of the equation, which is 0, equal to the derivative of the left side. Solve for \( y' \) to find the expression for the derivative.

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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, we differentiate both sides of the equation xy⁴ + x⁴y = 1 with respect to x, treating y as a function of x. This allows us to find the derivative y' without needing to solve for y explicitly.
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Product Rule
The product rule is a fundamental rule in calculus used to differentiate products of two functions. It states that if you have two functions u(x) and v(x), the derivative of their product is given by u'v + uv'. In the context of the given equation, we will apply the product rule to differentiate terms like xy⁴ and x⁴y, where both x and y are functions of x.
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Chain Rule
The chain rule is a method for differentiating composite functions. It states that if a variable y depends on u, which in turn depends on x, then the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x. In this problem, we will use the chain rule to differentiate terms involving y, as y is implicitly defined in terms of x.
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