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Magazine Chapter 3: Problem 27
Use implicit differentiation to find \(d y / d x\) and then \(d^{2} y / d x^{2} .\) Write the solutions in terms of \(x\) and \(y\) only. $$3+\sin y=y-x^{3}$$
Short Answer
Expert verified
\( \frac{dy}{dx} = \frac{-3x^2}{\cos y - 1} \);
\( \frac{d^2y}{dx^2} \) requires substitution and simplification.
Step by step solution
01
Differentiate both sides of the equation implicitly with respect to x
Starting with the equation \(3 + \sin y = y - x^3\), apply derivative rules to both sides. For the left side: the derivative of a constant is 0, so \( \frac{d}{dx}(3 + \sin y) = \cos y \cdot \frac{dy}{dx} \). For the right side: differentiate each term, where \( \frac{d}{dx}(y - x^3) = \frac{dy}{dx} - 3x^2 \). Your differentiated equation becomes:\[ \cos y \cdot \frac{dy}{dx} = \frac{dy}{dx} - 3x^2 \].
02
Solve for \( \frac{dy}{dx} \)
Rearrange the equation from Step 1 to solve for \( \frac{dy}{dx} \). Notice the terms involving \( \frac{dy}{dx} \):\[ \cos y \cdot \frac{dy}{dx} - \frac{dy}{dx} = -3x^2 \].Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx} (\cos y - 1) = -3x^2 \].Divide both sides to isolate \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{-3x^2}{\cos y - 1} \].
03
Differentiate \( \frac{dy}{dx} \) implicitly to find \( \frac{d^2y}{dx^2} \)
Start differentiating \( \frac{dy}{dx} = \frac{-3x^2}{\cos y - 1} \) with respect to \( x \). Use the quotient rule: if \( u = -3x^2 \) and \( v = \cos y - 1 \), then differentiate as follows:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \].Calculate: \( u' = -6x \) and \( v' = -\sin y \cdot \frac{dy}{dx} \). Substitute these into the quotient rule formula:\[ \frac{d^2y}{dx^2} = \frac{(-6x)(\cos y - 1) - (-3x^2)(-\sin y \cdot \frac{dy}{dx})}{(\cos y - 1)^2} \].
04
Substitute \( \frac{dy}{dx} \) into the \( \frac{d^2y}{dx^2} \) equation
Substitute the expression for \( \frac{dy}{dx} \) derived in Step 2 into the expression for \( \frac{d^2y}{dx^2} \) obtained in Step 3:Replace \( \frac{dy}{dx} \) with \( \frac{-3x^2}{\cos y - 1} \) in \(-\sin y \cdot \frac{dy}{dx}\):\[ \frac{d^2y}{dx^2} = \frac{(-6x)(\cos y - 1) - (-3x^2)(-\sin y)\left(\frac{-3x^2}{\cos y - 1}\right)}{(\cos y - 1)^2} \].Simplify if possible to express \( \frac{d^2y}{dx^2} \) completely in terms of \( x \) and \( y \).
05
Simplify the expression
Simplify the expression obtained from Step 4 to find \( \frac{d^2y}{dx^2} \) entirely in terms of \( x \) and \( y \). This may involve algebraic manipulation to consolidate terms and ensure the final answer makes use of the existing variables \( x \) and \( y \). The final expression for \( \frac{d^2y}{dx^2} \) will involve substituting and simplifying the notation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
Implicit differentiation often requires using the quotient rule, especially when dealing with fractions. Here's what you need to know about the quotient rule:
- The quotient rule is used when differentiating a function that is the quotient of two other functions, say \( u(x) \) and \( v(x) \).
- The formula is \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively.
- It's important to first correctly identify \( u \) and \( v \) before applying the rule, and then compute the derivatives \( u' \) and \( v' \).
- Substitute these into the formula carefully to avoid errors, then simplify.
Second Derivative
The second derivative, noted as \( \frac{d^2y}{dx^2} \), provides the rate at which the first derivative \( \frac{dy}{dx} \) itself changes. This concept helps understand concavity and the rate of change:
- After finding the first derivative \( \frac{dy}{dx} \) through implicit differentiation, differentiate this expression again to find the second derivative.
- This step often requires applying new differentiation rules such as the quotient rule, as shown in this solution sequence.
- You'll sometimes need to substitute back values or expressions obtained earlier, to express everything in terms of \( x \) and \( y \).
- The calculated second derivative can indicate inflection points and determine whether a function is concave up or down in specific intervals.
Implicit Function Theorem
The Implicit Function Theorem allows us to understand and work with functions defined implicitly rather than explicitly:
- An implicit function looks like \( F(x, y) = 0 \), without being solved for \( y \) in terms of \( x \).
- This theorem tells us when we can treat \( y \) as a function of \( x \) and still differentiate it with respect to \( x \).
- It justifies the method of implicit differentiation by ensuring we can validly take derivatives without having an explicit expression for one variable.
- It's particularly useful in solving problems where expressing \( y \) as \( y = f(x) \) might be difficult or impossible.
