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Find the implicit derivative $\frac{d}{dx}\left(y+x^2y=0\right)$

Used Formulas

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Solution

Formulas

Videos

Basic Derivatives

· Derivative of a Constant
$\frac{d}{dx}\left(c\right)=0$
· Derivative of the linear function
$\frac{d}{dx}\left(x\right)=1$
· Sum Rule for Differentiation
$\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}f\left(x\right) + \frac{d}{dx}g\left(x\right)$
· Product rule for derivatives
$\frac{d}{dx}\left(ab\right)=\frac{d}{dx}\left(a\right)b+a\frac{d}{dx}\left(b\right)$
· Power rule for derivatives
$\frac{d}{dx}\left(x^a\right)=ax^{\left(a-1\right)}$

Function Plot

Plotting: $y^{\prime}=\frac{-2xy}{1+x^2}$

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0
a
b
c
d
f
g
m
n
u
v
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x
y
z
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(◻)
+
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×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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

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$\frac{d}{dx}\left(x^2-y^2=25\right)$

Main Topic: Implicit Differentiation

Implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. For differentiating an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y(x) and then differentiate. Instead, one can differentiate R(x, y) with respect to x and y and then solve a linear equation in dy/dx for getting explicitly the derivative in terms of x and y.

Used Formulas

See formulas (5)

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