Final Answer
$y^{\prime}=\frac{-2x-y}{x-2y}$
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Step-by-step Solution
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1
Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable
$\frac{d}{dx}\left(x^2+xy-y^2\right)=\frac{d}{dx}\left(1\right)$
2
The derivative of the constant function ($1$) is equal to zero
$\frac{d}{dx}\left(x^2+xy-y^2\right)=0$
3
The derivative of a sum of two or more functions is the sum of the derivatives of each function
$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(-y^2\right)=0$
4
The derivative of a function multiplied by a constant ($-1$) is equal to the constant times the derivative of the function
$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(xy\right)-\frac{d}{dx}\left(y^2\right)=0$
5
Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=x$ and $g=y$
$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(x\right)y+x\frac{d}{dx}\left(y\right)-\frac{d}{dx}\left(y^2\right)=0$
Intermediate steps
6
The derivative of the linear function is equal to $1$
$\frac{d}{dx}\left(x^2\right)+y+x\frac{d}{dx}\left(y\right)-\frac{d}{dx}\left(y^2\right)=0$
Explain this step further
Intermediate steps
7
The derivative of the linear function is equal to $1$
$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-\frac{d}{dx}\left(y^2\right)=0$
Explain this step further
8
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-2y\frac{d}{dx}\left(y\right)=0$
Intermediate steps
9
The derivative of the linear function is equal to $1$
$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-2y\cdot y^{\prime}=0$
Explain this step further
Intermediate steps
10
The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$
$2x+y+xy^{\prime}-2y\cdot y^{\prime}=0$
Explain this step further
11
Group the terms of the equation by moving the terms that have the variable $y^{\prime}$ to the left side, and those that do not have it to the right side
$xy^{\prime}-2y\cdot y^{\prime}=-2x-y$
12
Factor the polynomial $xy^{\prime}-2y\cdot y^{\prime}$ by it's greatest common factor (GCF): $y^{\prime}$
$y^{\prime}\left(x-2y\right)=-2x-y$
13
Divide both sides of the equation by $x-2y$
$\frac{\left(x-2y\right)y^{\prime}}{x-2y}=\frac{-2x-y}{x-2y}$
14
Simplifying the quotients
$y^{\prime}=\frac{-2x-y}{x-2y}$
Final Answer$y^{\prime}=\frac{-2x-y}{x-2y}$