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)

검증된 단계별 안내

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 \(y' = \frac{dy}{dx}\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.

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)