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Magazine Chapter 3: Problem 25
Differentiate implicily to find \(d y / d x\). $$ \frac{x^{2} y+x y+1}{2 x+y}=1 \text { (Hint: Simplify first.) } $$
Short Answer
Expert verified
\( \frac{d y}{d x} = \frac{-2x - y }{ x^2 + x - 1} \)
Step by step solution
01
Simplify the Equation
First, simplify the given equation equation: \[ \frac{x^2 y + x y + 1}{2x + y} = 1 \]Multiply both sides of the equation by \[2x + y\]
02
Expand the Equation
Expand the equation: \[(x^2 y + x y + 1) = 1 * (2x + y)\]Simplify further:\[x^2 y + xy + 1 = 2x + y\]
03
Rearrange the Terms
Move all terms to one side of the equation to set it to 0:\[x^2 y + x y + 1 - 2x - y = 0 \]
04
Differentiate Implicitly
Differentiate both sides of the equation with respect to \(x\). Use the product rule for differentiation.The product rule states \[\frac{d}{dx}[u v] = u \frac{d v}{d x} + v \frac{d u}{d x}\] Differentiation of the given equation gives: \[ (x^2) \frac{d y}{d x} + y \frac{d (x^2)}{d x} + x \frac{d y}{d x} + y \frac{d x}{d x} +\frac{d}{dx}[ 1 - 2x - y] = 0 \]\ Simplifying the derivatives: \[ x^2 \frac{d y}{d x} + 2xy + x \frac{d y}{d x} + y -2 \frac{d x}{d x} - \frac{d y}{d x} =0 \]\ simplify to: \[ x^2 \frac{d y}{d x} + x \frac{d y}{d x} - \frac{d y}{d x}= -2x - y \]
05
Factor Out \(\frac{d y}{d x}\)
Since \( \frac{d y}{d x} \) is a common factor on the left-hand side, factor it out: \[\frac{d y }{d x} (x^2 + x - 1 ) = -2x - y\]
06
Solve for \(\frac{d y}{d x}\)
Divide both sides of the equation by \[ (x^2 + x - 1 ) \] to isolate \( \frac{d y}{d x}\): \[ \frac{d y}{d x} = \frac{-2x -y }{x^2 + x - 1} \]Hence, the implicit derivative is: \( \frac{d y}{d x} = \frac{-2x - y }{ x^2 + x - 1} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculus
Calculus is a branch of mathematics that studies how things change. It is split into two main areas: differential calculus and integral calculus. In simple terms, differential calculus focuses on rates of change, such as finding the slope of a curve. Integral calculus deals with accumulation of quantities, like finding the area under a curve.
Our focus here is on differential calculus, particularly implicit differentiation. This is useful in solving problems involving curves and slopes when the function is not explicitly given as y = f(x).
Our focus here is on differential calculus, particularly implicit differentiation. This is useful in solving problems involving curves and slopes when the function is not explicitly given as y = f(x).
Implicit Functions
Implicit functions are relationships between variables where the function is not isolated on one side of the equation. For example, in our exercise, the relationship between x and y is given as:
\ \[ \frac{x^{2} y + x y + 1}{2 x + y} = 1 \ \]
To solve for the derivative in these cases, we use implicit differentiation. This method treats both x and y as variables rather than isolating y beforehand. It's especially useful when isolating y is too complex or impossible.
In our problem, simplifying the equation first and differentiating implicitly allowed us to find \( \frac{d y}{d x} \) without explicitly solving for y.
\ \[ \frac{x^{2} y + x y + 1}{2 x + y} = 1 \ \]
To solve for the derivative in these cases, we use implicit differentiation. This method treats both x and y as variables rather than isolating y beforehand. It's especially useful when isolating y is too complex or impossible.
In our problem, simplifying the equation first and differentiating implicitly allowed us to find \( \frac{d y}{d x} \) without explicitly solving for y.
Derivative Rules
To differentiate a function, various derivative rules can be applied. Here are some of the key rules:
In our problem, we applied these rules to differentiate each term in the implicitly given equation. Specifically, the product rule played a crucial role.
- Power Rule: \( \frac{d}{dx} [x^n] = nx^{n-1} \)
- Constant Rule: \( \frac{d}{dx} [c] = 0 \) where c is a constant
- Sum Rule: \( \frac{d}{dx} [u + v] = \frac{d u}{d x} + \frac{d v}{d x} \)
- Product Rule: \( \frac{d}{dx} [uv] = u \frac{d v}{d x} + v \frac{d u}{d x} \)
In our problem, we applied these rules to differentiate each term in the implicitly given equation. Specifically, the product rule played a crucial role.
Product Rule
The product rule is essential in situations where we need to differentiate the product of two functions. The rule states:
\ \[ \frac{d}{dx} [uv] = u \frac{d v}{d x} + v \frac{d u}{d x} \ \]
In our exercise, we used the product rule to differentiate terms like \(x^2 y\) and \(xy\). For instance, to differentiate \(x^2 y\), we set:
Applying the product rule:
\ \[ \frac{d}{dx}[x^2 y] = x^2 \frac{d y}{d x} + 2xy \ \]
This step-by-step application of the product rule helped us successfully find the implicit derivative in the original problem.
\ \[ \frac{d}{dx} [uv] = u \frac{d v}{d x} + v \frac{d u}{d x} \ \]
In our exercise, we used the product rule to differentiate terms like \(x^2 y\) and \(xy\). For instance, to differentiate \(x^2 y\), we set:
- u = \(x^2 \)
- v = y
- \( \frac{d u}{d x} = 2x \)
- \( \frac{d v}{d x} = \frac{d y}{d x} \)
Applying the product rule:
\ \[ \frac{d}{dx}[x^2 y] = x^2 \frac{d y}{d x} + 2xy \ \]
This step-by-step application of the product rule helped us successfully find the implicit derivative in the original problem.
