Find the derivative of xy-1y^2+x^2=1

\frac{d}{dx}\left(x^2+xy-y^2=1\right)

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Answer

$2x+y=0$

Step by step solution

Problem

$\frac{d}{dx}\left(x^2+xy-y^2=1\right)$
1

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(-y^2+y\cdot x+x^2\right)=\frac{d}{dx}\left(1\right)$
2

The derivative of the constant function is equal to zero

$\frac{d}{dx}\left(-y^2+y\cdot x+x^2\right)=0$
3

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(-y^2\right)+\frac{d}{dx}\left(y\cdot x\right)+\frac{d}{dx}\left(x^2\right)=0$
4

The derivative of the constant function is equal to zero

$0+\frac{d}{dx}\left(y\cdot x\right)+\frac{d}{dx}\left(x^2\right)=0$
5

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$0+y\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(x^2\right)=0$
6

The derivative of the linear function is equal to $1$

$0+1y+\frac{d}{dx}\left(x^2\right)=0$
7

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$0+1y+2x=0$
8

Any expression multiplied by $1$ is equal to itself

$0+y+2x=0$
9

$x+0=x$, where $x$ is any expression

$2x+y=0$

Answer

$2x+y=0$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.21 seconds

Views:

171