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

## Step-by-step Solution

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sin
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asin
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asinh
acosh
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### Videos

$y^{\prime}=\frac{-2x-y}{x-2y}$
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## Step-by-step Solution

Problem to solve:

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

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

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

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

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

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

Any expression multiplied by $1$ is equal to itself

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

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

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

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

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

Any expression multiplied by $1$ is equal to itself

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

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

Any expression multiplied by $1$ is equal to itself

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

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

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

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

$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-1\cdot 2y^{\left(2-1\right)}\frac{d}{dx}\left(y\right)=0$

Subtract the values $2$ and $-1$

$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-1\cdot 2y^{1}\frac{d}{dx}\left(y\right)=0$

Multiply $-1$ times $2$

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

Any expression to the power of $1$ is equal to that same expression

$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-2y\frac{d}{dx}\left(y\right)=0$
8

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

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

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

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

Any expression multiplied by $1$ is equal to itself

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

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

Any expression multiplied by $1$ is equal to itself

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

$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-2\cdot 1yy^{\prime}=0$

Multiply $-2$ times $1$

$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-2yy^{\prime}=0$
9

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

$\frac{d}{dx}\left(x^2\right)+y+xy^{\prime}-2yy^{\prime}=0$

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

$2x^{\left(2-1\right)}+y+xy^{\prime}-2yy^{\prime}=0$

Subtract the values $2$ and $-1$

$2x^{1}+y+xy^{\prime}-2yy^{\prime}=0$

Any expression to the power of $1$ is equal to that same expression

$2x+y+xy^{\prime}-2yy^{\prime}=0$
10

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

$2x+y+xy^{\prime}-2yy^{\prime}=0$
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

$xy^{\prime}-2yy^{\prime}=-2x-y$
12

Factor the polynomial $xy^{\prime}-2yy^{\prime}$ by it's GCF: $y^{\prime}$

$y^{\prime}\left(x-2y\right)=-2x-y$
13

Divide both sides of the equation by $x-2y$

$y^{\prime}=\frac{-2x-y}{x-2y}$

$y^{\prime}=\frac{-2x-y}{x-2y}$
SnapXam A2

### beta Got another answer? Verify it!

Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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