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Magazine Chapter 3: Problem 5
Use implicit differentiation to find \(d y / d x.\) $$x^{2}(x-y)^{2}=x^{2}-y^{2}$$
Short Answer
Expert verified
\(\frac{dy}{dx} = \frac{-(2x - 2x(x-y)^2 - 2x^2(x-y))}{2x^2(x-y) + 2y}\)
Step by step solution
01
Differentiate Both Sides with Respect to x
We start by differentiating both sides of the given equation with respect to \(x\). The given equation is:\[x^{2}(x-y)^{2} = x^{2} - y^{2}\]Differentiating the left side with respect to \(x\), we use the product rule and chain rule:\[\frac{d}{dx}[x^{2}(x-y)^{2}] = \frac{d}{dx}[x^2] \cdot (x-y)^2 + x^2 \cdot \frac{d}{dx}[(x-y)^2]\] This becomes:\[2x(x-y)^2 + x^2 \cdot 2(x-y)(1 - \frac{dy}{dx})\]For the right side, applying direct differentiation is straightforward:\[\frac{d}{dx}[x^2 - y^2] = 2x - 2y \cdot \frac{dy}{dx}\]
02
Set the Derivatives Equal
Equate the derivatives obtained from differentiating both sides:\[2x(x-y)^2 + 2x^2(x-y)(1-\frac{dy}{dx}) = 2x - 2y \cdot \frac{dy}{dx}\]This is our equation to solve for \(\frac{dy}{dx}\).
03
Simplify the Equation
First, distribute and simplify the left side:\[2x(x-y)^2 + 2x^2(x-y) - 2x^2(x-y)\frac{dy}{dx} = 2x - 2y\frac{dy}{dx}\]Combine like terms:\[2x(x-y)^2 + 2x^2(x-y) = 2x - 2x^2(x-y)\frac{dy}{dx} - 2y\frac{dy}{dx}\]
04
Collect Terms Involving dy/dx
Move terms involving \(\frac{dy}{dx}\) to one side:\[2x - 2x(x-y)^2 - 2x^2(x-y) = -2x^2(x-y)\frac{dy}{dx} - 2y\frac{dy}{dx}\]
05
Solve for dy/dx
Factor \(\frac{dy}{dx}\) out:\[-2x^2(x-y)\frac{dy}{dx} - 2y\frac{dy}{dx} = 2x - 2x(x-y)^2 - 2x^2(x-y)\]Re-express terms to isolate \(\frac{dy}{dx}\):\[\frac{dy}{dx}(2x^2(x-y) + 2y) = -(2x - 2x(x-y)^2 - 2x^2(x-y))\]Finally, solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{-(2x - 2x(x-y)^2 - 2x^2(x-y))}{2x^2(x-y) + 2y}\]
06
Simplify the Expression for dy/dx
Simplify the expression further if possible:In many cases, the expression may simplify depending on the context or specific values. However, in this case, it represents the rate of change dy/dx based on the differentiated equation.Therefore, the simplified expression for \(\frac{dy}{dx}\) is:\[\frac{dy}{dx} = \frac{-(2x - 2x(x-y)^2 - 2x^2(x-y))}{2x^2(x-y) + 2y}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The **Product Rule** is a fundamental concept in calculus used to differentiate expressions where two functions are being multiplied. If you have a product of functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative is:\[\frac{d}{dx}[u(x) \, v(x)] = u'(x)v(x) + u(x)v'(x)\]
In our exercise, the function \( u(x) = x^2 \) and \( v(x) = (x-y)^2 \) are multiplied together. Applying the product rule here helps us break down this multiplication into:
In our exercise, the function \( u(x) = x^2 \) and \( v(x) = (x-y)^2 \) are multiplied together. Applying the product rule here helps us break down this multiplication into:
- Take the derivative of the first function \( x^2 \) and multiply it by the second function \( (x-y)^2 \).
- Keep the first function \( x^2 \) the same and multiply it by the derivative of the second function \((x-y)^2\).
Chain Rule
The **Chain Rule** is essential when differentiating compositions of functions. It allows you to "chain" the differentiation of multiple functions together. The basic principle is: if you have a composite function \( y = f(g(x)) \), then the derivative is given by:\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]
In the exercise, we have \((x-y)^2\) which is a function within a function. It's a composition of the outer function \( u^2 \) and the inner function \( u = (x-y) \). This calls for the application of the chain rule:
In the exercise, we have \((x-y)^2\) which is a function within a function. It's a composition of the outer function \( u^2 \) and the inner function \( u = (x-y) \). This calls for the application of the chain rule:
- First, differentiate the outer function \( u^2 \) to get \( 2u \).
- Then, differentiate the inner function \( x-y \), giving \( 1 - \frac{dy}{dx} \) since \( x \) differentiates to 1 and \( y \) to \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \).
Dy/Dx Calculation
The calculation of \( \frac{dy}{dx} \) via implicit differentiation involves first differentiating both sides of the given equation with respect to \( x \). After differentiating, as seen in the step-by-step guide, we gather all the terms that include \( \frac{dy}{dx} \) to one side, creating an equation that can be solved for \( \frac{dy}{dx} \).
Here's a recap of the process:
This systematic approach to implicit differentiation allows you to solve for derivatives where \( y \) isn't isolated, showcasing the strength of calculus tools for tackling complex problems.
Here's a recap of the process:
- Differentiating each side of the equation separately using proper rules and replacing derivatives accordingly, such as using the product and chain rules as needed.
- Rearranging the equation to move all terms involving \( \frac{dy}{dx} \) to one side while moving other terms to the opposite side.
- Factoring out \( \frac{dy}{dx} \) from these terms to isolate it effectively.
- Solving the resulting equation for \( \frac{dy}{dx} \).
This systematic approach to implicit differentiation allows you to solve for derivatives where \( y \) isn't isolated, showcasing the strength of calculus tools for tackling complex problems.
