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Find the implicit derivative $\frac{d}{dx}\left(ax^2-3b^2xy+cy^2=1\right)$

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Solution

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Final answer to the problem

$2ax-3b^2\left(y+xy^{\prime}\right)+2cy\cdot y^{\prime}=0$
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Step-by-step Solution

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  • Find the derivative using the definition
  • Find the derivative using the product rule
  • Find the derivative using the quotient rule
  • Find the derivative using logarithmic differentiation
  • Find the derivative
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
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1

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

$\frac{d}{dx}\left(ax^2-3b^2xy+cy^2\right)=\frac{d}{dx}\left(1\right)$

Learn how to solve implicit differentiation problems step by step online.

$\frac{d}{dx}\left(ax^2-3b^2xy+cy^2\right)=\frac{d}{dx}\left(1\right)$

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Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(ax^2-3b^2xycy^2=1). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The derivative of the constant function (1) is equal to zero. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function.

Final answer to the problem

$2ax-3b^2\left(y+xy^{\prime}\right)+2cy\cdot y^{\prime}=0$

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Plotting: $2ax-3b^2\left(y+xy^{\prime}\right)+2cy\cdot y^{\prime}=0$

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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
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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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

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