Derive the function x^3+y^3=xy*6-1 with respect to x

\frac{d}{dx}\left(x^3+y^3=6x\cdot y-1\right)

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Answer

$3x^{2}=6y$

Step by step solution

Problem

$\frac{d}{dx}\left(x^3+y^3=6x\cdot y-1\right)$
1

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

$\frac{d}{dx}\left(y^3+x^3\right)=\frac{d}{dx}\left(6y\cdot x-1\right)$
2

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(y^3\right)+\frac{d}{dx}\left(x^3\right)=\frac{d}{dx}\left(-1\right)+\frac{d}{dx}\left(6y\cdot x\right)$
3

The derivative of the constant function is equal to zero

$0+\frac{d}{dx}\left(x^3\right)=0+\frac{d}{dx}\left(6y\cdot x\right)$
4

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=6x$ and $g=y$

$0+\frac{d}{dx}\left(x^3\right)=0+6x\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(6x\right)$
5

The derivative of the constant function is equal to zero

$0+\frac{d}{dx}\left(x^3\right)=0+0\cdot 6x+y\frac{d}{dx}\left(6x\right)$
6

Any expression multiplied by $0$ is equal to $0$

$0+\frac{d}{dx}\left(x^3\right)=0+0+y\frac{d}{dx}\left(6x\right)$
7

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$0+\frac{d}{dx}\left(x^3\right)=0+0+6y\frac{d}{dx}\left(x\right)$
8

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

$0+\frac{d}{dx}\left(x^3\right)=0+0+1\cdot 6y$
9

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

$0+3x^{2}=0+0+1\cdot 6y$
10

Add the values $0$ and $0$

$0+3x^{2}=6y$
11

$x+0=x$, where $x$ is any expression

$3x^{2}=6y$

Answer

$3x^{2}=6y$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.2 seconds

Views:

78