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# Find the implicit derivative $\frac{d}{dx}\left(x^3-y^3=x+8\right)$

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##  Final answer to the problem

$y^{\prime}=\frac{1-3x^{2}}{-3y^{2}}$
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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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Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(x^3-y^3\right)=\frac{d}{dx}\left(x+8\right)$

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

$\frac{d}{dx}\left(x^3-y^3\right)=\frac{d}{dx}\left(x+8\right)$

Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(x^3-y^3=x+8). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of the constant function (8) is equal to zero.

##  Final answer to the problem

$y^{\prime}=\frac{1-3x^{2}}{-3y^{2}}$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

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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

###  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.

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