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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.6.51
Chapter 7, Problem 7.6.51

For problems 49–52 use implicit differentiation to find dy/dx at the given point P.
51. y arccos(xy) = -3√2/4 π; P(1/2, -√2)

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Start with the given equation: \(y \arccos(xy) = -\frac{3\sqrt{2}}{4} \pi\).
Differentiate both sides of the equation with respect to \(x\) using implicit differentiation. Remember that \(y\) is a function of \(x\), so apply the product rule to \(y \arccos(xy)\) and the chain rule to \(\arccos(xy)\).
For the left side, apply the product rule: \(\frac{d}{dx}[y \arccos(xy)] = y' \arccos(xy) + y \frac{d}{dx}[\arccos(xy)]\). Then, differentiate \(\arccos(xy)\) using the chain rule: \(\frac{d}{dx}[\arccos(u)] = -\frac{1}{\sqrt{1-u^2}} \cdot u'\), where \(u = xy\).
Calculate \(u' = \frac{d}{dx}[xy]\) using the product rule: \(u' = y + x y'\). Substitute this back into the derivative of \(\arccos(xy)\).
After differentiating, substitute the point \(P\left(\frac{1}{2}, -\sqrt{2}\right)\) into the resulting expression and solve for \(y' = \frac{dy}{dx}\).

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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 find the derivative of a function when y is defined implicitly in terms of x, rather than explicitly. It involves differentiating both sides of the equation with respect to x, treating y as a function of x, and then solving for dy/dx.
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Finding The Implicit Derivative

Derivative of Inverse Trigonometric Functions

The derivative of inverse trigonometric functions, such as arccos(u), is essential when differentiating expressions involving these functions. For arccos(u), the derivative is -1 / √(1 - u²) times the derivative of u, which must be applied using the chain rule.
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Derivatives of Other Inverse Trigonometric Functions

Chain Rule

The chain rule is used to differentiate composite functions, where one function is inside another. When differentiating expressions like arccos(xy), you differentiate the outer function with respect to the inner function and multiply by the derivative of the inner function, applying product and implicit differentiation as needed.
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Intro to the Chain Rule
Related Practice
Textbook Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.

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L’Hôpital’s Rule

Find the limits in Exercises 103–110.

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Textbook Question

In Exercises 27–32, find dy/dx.

e^(2x)=sin(x+3y)

Textbook Question

In Exercises 1–4, solve for t.

e^(sqrt(t)) = x^2