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\frac{d}{dx}\left(x^3-3xy+y^3=1\right)

Find the derivative of xy*-3+y^3+x^3=1

Answer

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

Step-by-step explanation

Problem

$\frac{d}{dx}\left(x^3-3xy+y^3=1\right)$
1

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

$\frac{d}{dx}\left(y^3-3y\cdot x+x^3\right)=\frac{d}{dx}\left(1\right)$

Unlock this step-by-step solution!

Answer

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

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$\frac{d}{dx}\left(x^3-3xy+y^3=1\right)$

Main topic:

Differential calculus

Used formulas:

3. See formulas

Time to solve it:

~ 0.73 seconds