/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Find \(\frac{d y}{d x}\) and \(\... [FREE SOLUTION] | 91Ó°ÊÓ

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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.
In the problem, we utilize the Quotient Rule while differentiating the expression \( \frac{2-y}{x+3} \) to find \( \frac{d^2 y}{dx^2} \). Here:
  • \( u = 2 - y \).
  • \( v = x + 3 \).
  • The derivative \( u' \) equals \(-\frac{dy}{dx} \), as derived from the chain rule.
  • \( v' = 1 \), since the derivative of \( x+3 \) is straightforward.
By applying the Quotient Rule, you can effectively differentiate functions of this form. Remember to substitute back any expressions like \( \frac{dy}{dx} \) with their calculated values to simplify results.
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.
Looking at our original expression \( xy = 2x - 3y \), we employed the Product Rule to differentiate the term \( xy \) explicitly:
  • 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 \( xy' + x\frac{dy}{dx} \).
This approach simplifies the differentiation of products in implicit functions, essential in our exercise where each term factors into the re-arranged equation.
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 \( x \).
In the current exercise, the Chain Rule is used when differentiating the term involving \( y \). Because \( y \) is implicitly related to \( x \), when you differentiate such terms like \( 3y \), you must include the derivative of \( y \), which is \( \frac{dy}{dx} \). The derivative becomes \( 3 \cdot \frac{dy}{dx} \) applying the Chain Rule to account for the dependency of \( y \) on \( x \). This step is key when differentiating expressions in implicit differentiation, ensuring all variables related indirectly to \( x \) are appropriately differentiated.

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Most popular questions from this chapter

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Given that \(f(x)=\arcsin x+\arccos x,\) find \(f^{\prime}(x)\). What can you conclude about the function \(f ?\)

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