Find the derivative of y+xy^3*3+x^2=(1-1x)/x

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

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Answer

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^{2}}-\frac{1}{x}$

Step by step solution

Problem

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

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

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

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

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

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

$\frac{d}{dx}\left(3xy^3\right)+\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(x^2\right)=\frac{1\left(-1\right)\left(1-x\right)+x\left(\frac{d}{dx}\left(-x\right)+\frac{d}{dx}\left(1\right)\right)}{x^2}$
5

The derivative of the constant function is equal to zero

$\frac{d}{dx}\left(3xy^3\right)+0+\frac{d}{dx}\left(x^2\right)=\frac{1\left(-1\right)\left(1-x\right)+x\left(\frac{d}{dx}\left(-x\right)+0\right)}{x^2}$
6

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

$\frac{d}{dx}\left(3xy^3\right)+0+\frac{d}{dx}\left(x^2\right)=\frac{1\left(-1\right)\left(1-x\right)+x\left(0-\frac{d}{dx}\left(x\right)\right)}{x^2}$
7

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

$\frac{d}{dx}\left(3xy^3\right)+0+\frac{d}{dx}\left(x^2\right)=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
8

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=3x$ and $g=y^3$

$0+\frac{d}{dx}\left(x^2\right)+3x\frac{d}{dx}\left(y^3\right)+y^3\frac{d}{dx}\left(3x\right)=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
9

The derivative of the constant function is equal to zero

$0+\frac{d}{dx}\left(x^2\right)+0\cdot 3x+y^3\frac{d}{dx}\left(3x\right)=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
10

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

$0+\frac{d}{dx}\left(x^2\right)+0+y^3\frac{d}{dx}\left(3x\right)=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
11

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

$0+\frac{d}{dx}\left(x^2\right)+0+3y^3\frac{d}{dx}\left(x\right)=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
12

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

$0+\frac{d}{dx}\left(x^2\right)+0+1\cdot 3y^3=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
13

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

$0+2x+0+1\cdot 3y^3=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
14

Add the values $0$ and $0$

$2x+1\cdot 3y^3+0=\frac{1\left(-1\right)\left(1-x\right)+\left(1\left(-1\right)+0\right)x}{x^2}$
15

Multiply $3$ times $1$

$2x+3y^3+0=\frac{\left(0-1\right)x-\left(1-x\right)}{x^2}$
16

Subtract the values $0$ and $-1$

$2x+3y^3+0=\frac{-x-\left(1-x\right)}{x^2}$
17

Split the fraction $\frac{-x+-\left(1-x\right)}{x^2}$ in two terms with same denominator

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^2}+\frac{-x}{x^2}$
18

Simplifying the fraction by $x$

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^2}+\frac{-1}{x}$
19

Unifying fractions

$2x+3y^3+0=\frac{-x^2-x\left(1-x\right)}{x^{3}}$
20

Split the fraction $\frac{-x^2+-x\left(1-x\right)}{x^{3}}$ in two terms with same denominator

$2x+3y^3+0=\frac{-x\left(1-x\right)}{x^{3}}+\frac{-x^2}{x^{3}}$
21

Simplifying the fraction by $x$

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^{2}}+\frac{-x^2}{x^{3}}$
22

Simplifying the fraction by $x$

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^{2}}-x^{\left(2-3\right)}$
23

Subtract the values $2$ and $-3$

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^{2}}-x^{-1}$
24

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^{2}}-\frac{1}{x}$

Answer

$2x+3y^3+0=\frac{-\left(1-x\right)}{x^{2}}-\frac{1}{x}$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.35 seconds

Views:

126