Chapter 15: Problem 21
Find \(\frac{d y}{d x}\) and \(\frac{d^{2} y}{d x^{2}}\) for the given equation. $$x y=2 x-3 y$$
Short Answer
Expert verified
\(\frac{dy}{dx} = \frac{2-y}{x+3}\) and \(\frac{d^2 y}{dx^2} = \frac{2y-4}{(x+3)^2}\).
Step by step solution
01
Simplify the Equation
Start with the given equation: \(xy = 2x - 3y\). Rearrange it to bring all terms to one side: \(xy + 3y - 2x = 0\). This simplifies the implicit differentiation process in the upcoming steps.
02
Differentiate Implicitly with Respect to x
Differentiate each term of the equation \(xy + 3y - 2x = 0\) with respect to \(x\).- The derivative of \(xy\) involves the product rule: \((1)(y) + (x)\frac{dy}{dx} = y + x\frac{dy}{dx}\).- The derivative of \(3y\) is \(3\frac{dy}{dx}\), using the chain rule.- The derivative of \(-2x\) is \(-2\).So, the differentiated equation is: \(y + x\frac{dy}{dx} + 3\frac{dy}{dx} - 2 = 0\).
03
Solve for \(\frac{dy}{dx}\)
Rearrange the differentiated equation from Step 2 to solve for \(\frac{dy}{dx}\):1. Combine like terms: \(x\frac{dy}{dx} + 3\frac{dy}{dx} = -y + 2\).2. Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(x + 3) = 2 - y \).3. Isolate \(\frac{dy}{dx}\) by dividing both sides by \(x + 3\): \[\frac{dy}{dx} = \frac{2 - y}{x + 3}\].
04
Differentiate \(\frac{dy}{dx}\) with Respect to x
To find \(\frac{d^2 y}{dx^2}\), differentiate \(\frac{dy}{dx} = \frac{2 - y}{x + 3}\). Use the quotient rule, which states \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\), where \(u = 2 - y\) and \(v = x + 3\).- \(u' = -\frac{dy}{dx}\) (since the derivative of \(-y\) is \(-\frac{dy}{dx}\)).- \(v' = 1\).Apply the quotient rule:\[\frac{d}{dx}\left(\frac{2-y}{x+3}\right) = \frac{(-\frac{dy}{dx})(x+3) - (2-y)(1)}{(x+3)^2}\],which simplifies to:\[\frac{d^2 y}{dx^2} = \frac{-\frac{dy}{dx}(x+3) - (2-y)}{(x+3)^2}\].
05
Substitute \(\frac{dy}{dx}\) in the Second Derivative Expression
Replace \(\frac{dy}{dx}\) in the \(\frac{d^2 y}{dx^2}\) expression from Step 4 with its value from Step 3:1. Substitute \(\frac{dy}{dx}\) as \(\frac{2-y}{x+3}\): \[\frac{d^2 y}{dx^2} = \frac{-\left(\frac{2-y}{x+3}\right)(x+3) - (2-y)}{(x+3)^2}\].2. Simplify: \(-\left(2-y\right) - (2-y) = -2 + y - 2 + y = 2y - 4\).3. The simplified second derivative is: \[\frac{d^2 y}{dx^2} = \frac{2y - 4}{(x+3)^2}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The Quotient Rule is a handy tool for differentiating functions expressed as fractions. It helps when you have one function divided by another, like \( \frac{u}{v} \). The rule is summarized as: \[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] Where:
- \( u \) is the numerator function.
- \( v \) is the denominator function.
- \( u' \) and \( v' \) are their respective derivatives.
- \( u = 2 - y \).
- \( v = x + 3 \).
- The derivative \( u' \) equals \(-\frac{dy}{dx} \), as derived from the chain rule.
- \( v' = 1 \), since the derivative of \( x+3 \) is straightforward.
Product Rule
The Product Rule is a method for differentiating products of two functions. If you have a product \( uv \), where \( u \) and \( v \) are both functions of \( x \), the rule states: \[\frac{d}{dx}(uv) = u'v + uv'\] Where:
- \( u \) and \( v \) are the two functions you are differentiating.
- \( u' \) and \( v' \) are their derivatives.
- With \( x \) and \( y \) treated as the two functions, \( u' = 1 \) since the derivative of \( x \) is 1.
- \( v \), the derivative of \( y \), is \( \frac{dy}{dx} \).
- Applying the rule, it becomes \( xy' + x\frac{dy}{dx} \).
Chain Rule
The Chain Rule is crucial when dealing with composite functions. It allows you to find the derivative of functions nested within one another. The formula is as follows: \[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\] Where:
- \( f \) is the outer function, applied to \( g(x) \).
- \( g \) is the inner function dependent on \( x \).