# Derive the function 4x^2+xy*2+y^3=x+6y with respect to x

## \frac{d}{dx}\left(y^3+4x^2+2x\cdot y=x+6y\right)

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$8x+2y+0=1$

## Step by step solution

Problem

$\frac{d}{dx}\left(y^3+4x^2+2x\cdot y=x+6y\right)$
1

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

$\frac{d}{dx}\left(2y\cdot x+4x^2+y^3\right)=\frac{d}{dx}\left(6y+x\right)$
2

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

$\frac{d}{dx}\left(2y\cdot x\right)+\frac{d}{dx}\left(4x^2\right)+\frac{d}{dx}\left(y^3\right)=\frac{d}{dx}\left(6y\right)+\frac{d}{dx}\left(x\right)$
3

The derivative of the constant function is equal to zero

$\frac{d}{dx}\left(2y\cdot x\right)+\frac{d}{dx}\left(4x^2\right)+0=0+\frac{d}{dx}\left(x\right)$
4

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

$\frac{d}{dx}\left(2y\cdot x\right)+\frac{d}{dx}\left(4x^2\right)+0=0+1$
5

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

$\frac{d}{dx}\left(2y\cdot x\right)+4\frac{d}{dx}\left(x^2\right)+0=0+1$
6

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=2x$ and $g=y$

$4\frac{d}{dx}\left(x^2\right)+0+2x\frac{d}{dx}\left(y\right)+y\frac{d}{dx}\left(2x\right)=0+1$
7

The derivative of the constant function is equal to zero

$4\frac{d}{dx}\left(x^2\right)+0+0\cdot 2x+y\frac{d}{dx}\left(2x\right)=0+1$
8

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

$4\frac{d}{dx}\left(x^2\right)+0+0+y\frac{d}{dx}\left(2x\right)=0+1$
9

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

$4\frac{d}{dx}\left(x^2\right)+0+0+2y\frac{d}{dx}\left(x\right)=0+1$
10

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

$4\frac{d}{dx}\left(x^2\right)+0+0+1\cdot 2y=0+1$
11

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

$4\cdot 2x+0+0+1\cdot 2y=0+1$
12

Add the values $0$ and $0$

$4\cdot 2x+1\cdot 2y+0=1$
13

Multiply $2$ times $1$

$8x+2y+0=1$

$8x+2y+0=1$

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### Main topic:

Differential calculus

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