Textbook Question
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
cos y = x; (0, π/2)
4. Applications of Derivatives
Implicit Differentiation
4:39 minutesProblem 3.8.23a
Textbook Question
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
³√x+³√y⁴ = 2;(1,1)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. The equation is \( \sqrt[3]{x} + \sqrt[3]{y^4} = 2 \). Remember that \( y \) is a function of \( x \), so when differentiating terms involving \( y \), use implicit differentiation.
Differentiate \( \sqrt[3]{x} \) with respect to \( x \). This is \( \frac{d}{dx}(x^{1/3}) \), which results in \( \frac{1}{3}x^{-2/3} \).
Differentiate \( \sqrt[3]{y^4} \) with respect to \( x \). This requires the chain rule: \( \frac{d}{dx}(y^4)^{1/3} = \frac{1}{3}(y^4)^{-2/3} \cdot 4y^3 \cdot \frac{dy}{dx} \).
Set the derivative of the left side equal to the derivative of the right side. Since the derivative of a constant (2) is 0, you have: \( \frac{1}{3}x^{-2/3} + \frac{4}{3}y^{3}(y^4)^{-2/3} \cdot \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. This involves algebraic manipulation to express \( \frac{dy}{dx} \) in terms of \( x \) and \( y \).
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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. Instead of solving for y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This method allows us to find the derivative dy/dx without isolating y.
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Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When applying implicit differentiation, the chain rule is essential for differentiating terms involving y, as we must multiply by dy/dx when differentiating y with respect to x. This ensures that we account for the relationship between x and y in the differentiation process.
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Evaluating Derivatives at a Point
After finding the derivative dy/dx using implicit differentiation, we often need to evaluate it at a specific point, such as (1,1) in this case. This involves substituting the x and y values into the derived expression for dy/dx to find the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific coordinates.
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