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Magazine Chapter 3: Problem 1
Use implicit differentiation to find \(d y / d x\) in Exercises \(1-14\) \begin{equation} x^{2} y+x y^{2}=6 \end{equation}
Short Answer
Expert verified
\(\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}\).
Step by step solution
01
Differentiate both sides w.r.t. x
Start by differentiating both sides of the equation \(x^2 y + xy^2 = 6\) with respect to \(x\). Use the product rule where needed. The differentiation of a constant (6) with respect to \(x\) is 0.
02
Use the product rule
For the first term \(x^2 y\), use the product rule: \(d(u v) = u'v + uv'\).Thus, differentiate \(x^2 y\) to get\[\frac{d}{dx}(x^2 y) = 2xy + x^2 \frac{dy}{dx}.\]
03
Differentiate the second term
For the second term \(xy^2\), again use the product rule: \(uv' + u'v\). So,\[\frac{d}{dx}(x y^2) = y^2 + 2xy \frac{dy}{dx}.\]
04
Set up the equation
Combine the differentiated terms:\[2xy + x^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 0.\]
05
Simplify and group like terms
Move all terms involving \(\frac{dy}{dx}\) to one side of the equation:\[x^2 \frac{dy}{dx} + 2xy \frac{dy}{dx} = -2xy - y^2.\]
06
Factor and solve for \(\frac{dy}{dx}\)
Factor \(\frac{dy}{dx}\) out of the left side:\[\frac{dy}{dx}(x^2 + 2xy) = -2xy - y^2.\]Now, solve for \(\frac{dy}{dx}\) by dividing both sides by \(x^2 + 2xy\):\[\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}.\]
07
Simplify the derivative further if possible
The expression for \(\frac{dy}{dx}\) is \[\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}.\]Check if there's any common factor to simplify further. To maintain the relationship derived, this is the most simplified form.
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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 key concept in calculus. It is especially important when you need to differentiate expressions where two functions multiply each other. Here's an easy way to remember the product rule:
- When you have two functions, say \( u \) and \( v \), their derivative is given by \( d(uv) = u'v + uv' \).
- This means you differentiate one function while leaving the other unchanged, then switch roles.
- For \( x^2 y \), use the rule: the derivative is \( 2xy + x^2 \frac{dy}{dx}\).
- For \( xy^2 \), it becomes \( y^2 + 2xy \frac{dy}{dx} \).
Derivative
The derivative is a core tool in calculus, acting like a microscope that lets us zoom in on functions to understand how they change. Fundamentally, it measures the rate of change: how quickly or slowly a function's value is changing.
This is why, when we differentiate \( y \), we attach \( \frac{dy}{dx} \) to show it's a derivative with respect to \( x \). This gives us a comprehensive view of the function's behavior.
- When differentiating with respect to \( x \), you're looking at how the function changes as \( x \) changes.
- To find a derivative, you systematically apply rules like the power rule \((d/dx) x^n = n x^{n-1}\) and the product rule.
This is why, when we differentiate \( y \), we attach \( \frac{dy}{dx} \) to show it's a derivative with respect to \( x \). This gives us a comprehensive view of the function's behavior.
Calculus
Calculus is the mathematical study of change and motion. It's a vast field but one that's essential for understanding how things evolve over time and space. Two of the fundamental concepts in calculus are differentiation and integration.
\( \frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy} \), calculus lets us express relationships between the variables \( x \) and \( y \) and understand how these relationships evolve.
This makes calculus a powerful tool to decode complex real-world phenomena, from physics to economics.
- Differentiation, part of what we're doing in this exercise, allows us to find the slope of the tangent line to a curve at any point, effectively telling us the rate of change.
- Implicit differentiation, as we've applied here, helps us tackle equations where \( y \) is not isolated but interwoven with \( x \).
\( \frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy} \), calculus lets us express relationships between the variables \( x \) and \( y \) and understand how these relationships evolve.
This makes calculus a powerful tool to decode complex real-world phenomena, from physics to economics.
