Find the derivative of y=-12/(x^2)+2/(3x^3)+3/x

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

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Answer

$0=\frac{4}{x^{3}}+\left(\frac{-6}{x^{2}}-27\right)\frac{\frac{1}{9}}{x^{4}}$

Step by step solution

Problem

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

Taking out the constant $3$ from the fraction's denominator

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

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

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

The derivative of the constant function is equal to zero

$0=\frac{d}{dx}\left(\frac{\frac{1}{3}\cdot\frac{2}{x^3}}{3}-\frac{2}{x^2}+\frac{3}{x}\right)$
4

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

$0=\frac{d}{dx}\left(\frac{\frac{1}{3}\cdot\frac{2}{x^3}}{3}\right)+\frac{d}{dx}\left(-\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
5

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

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

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$0=\frac{3\frac{d}{dx}\left(\frac{1}{3}\cdot\frac{2}{x^3}\right)-\frac{1}{3}\cdot\frac{2}{x^3}\cdot\frac{d}{dx}\left(3\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
7

The derivative of the constant function is equal to zero

$0=\frac{0\left(-\frac{1}{3}\right)\left(\frac{2}{x^3}\right)+3\frac{d}{dx}\left(\frac{1}{3}\cdot\frac{2}{x^3}\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
8

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

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

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

$0=\frac{0+3\cdot \frac{1}{3}\cdot\frac{d}{dx}\left(\frac{2}{x^3}\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
10

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{x^3\frac{d}{dx}\left(2\right)-2\frac{d}{dx}\left(x^3\right)}{\left(x^3\right)^2}\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
11

The derivative of the constant function is equal to zero

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0x^3-2\frac{d}{dx}\left(x^3\right)}{\left(x^3\right)^2}\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
12

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

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\frac{d}{dx}\left(x^3\right)}{\left(x^3\right)^2}\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
13

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

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{d}{dx}\left(\frac{2}{x^2}\right)+\frac{d}{dx}\left(\frac{3}{x}\right)$
14

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{x^2\frac{d}{dx}\left(2\right)-2\frac{d}{dx}\left(x^2\right)}{\left(x^2\right)^2}+\frac{d}{dx}\left(\frac{3}{x}\right)$
15

The derivative of the constant function is equal to zero

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0x^2-2\frac{d}{dx}\left(x^2\right)}{\left(x^2\right)^2}+\frac{d}{dx}\left(\frac{3}{x}\right)$
16

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

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0-2\frac{d}{dx}\left(x^2\right)}{\left(x^2\right)^2}+\frac{d}{dx}\left(\frac{3}{x}\right)$
17

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

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0-2\cdot 2x}{\left(x^2\right)^2}+\frac{d}{dx}\left(\frac{3}{x}\right)$
18

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0-2\cdot 2x}{\left(x^2\right)^2}+\frac{x\frac{d}{dx}\left(3\right)-3\frac{d}{dx}\left(x\right)}{x^2}$
19

The derivative of the constant function is equal to zero

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0-2\cdot 2x}{\left(x^2\right)^2}+\frac{0x-3\frac{d}{dx}\left(x\right)}{x^2}$
20

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

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0-2\cdot 2x}{\left(x^2\right)^2}+\frac{0-3\frac{d}{dx}\left(x\right)}{x^2}$
21

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

$0=\frac{0+3\cdot \frac{1}{3}\left(\frac{0-2\cdot 3x^{2}}{\left(x^3\right)^2}\right)}{9}-\frac{0-2\cdot 2x}{\left(x^2\right)^2}+\frac{1\left(-3\right)+0}{x^2}$
22

Multiply $-3$ times $1$

$0=\frac{0+1\frac{0-6x^{2}}{\left(x^3\right)^2}}{9}-\frac{0-4x}{\left(x^2\right)^2}+\frac{0-3}{x^2}$
23

Subtract the values $0$ and $-3$

$0=\frac{0+1\frac{0-6x^{2}}{\left(x^3\right)^2}}{9}-\frac{0-4x}{\left(x^2\right)^2}+\frac{-3}{x^2}$
24

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

$0=\frac{\frac{-6x^{2}}{\left(x^3\right)^2}}{9}-\frac{-4x}{\left(x^2\right)^2}+\frac{-3}{x^2}$
25

Applying the power of a power property

$0=\frac{\frac{-6x^{2}}{x^{6}}}{9}-\frac{-4x}{x^{4}}+\frac{-3}{x^2}$
26

Simplifying the fraction by $x$

$0=\frac{\frac{-6x^{2}}{x^{6}}}{9}-\frac{-4}{x^{3}}+\frac{-3}{x^2}$
27

Simplifying the fraction by $x$

$0=\frac{-6x^{\left(2-6\right)}}{9}-\frac{-4}{x^{3}}+\frac{-3}{x^2}$
28

Subtract the values $2$ and $-6$

$0=\frac{-6x^{-4}}{9}-\frac{-4}{x^{3}}+\frac{-3}{x^2}$
29

Unifying fractions

$0=\frac{4}{x^{3}}+\frac{-27-6x^{-2}}{9x^2}$
30

Taking out the constant $9$ from the fraction's denominator

$0=\frac{4}{x^{3}}+\frac{\frac{1}{9}\cdot\frac{-27-6x^{-2}}{x^2}}{x^2}$
31

Multiplying the fraction and term

$0=\frac{4}{x^{3}}+\frac{\frac{\frac{1}{9}\left(-27-6x^{-2}\right)}{x^2}}{x^2}$
32

Simplifying the fraction

$0=\frac{4}{x^{3}}+\frac{\frac{1}{9}\left(-27-6x^{-2}\right)}{x^2x^2}$
33

When multiplying exponents with same base we can add the exponents

$0=\frac{4}{x^{3}}+\frac{\frac{1}{9}\left(-27-6x^{-2}\right)}{x^{4}}$
34

Unifying fractions

$0=\frac{4x^{4}+\frac{1}{9}x^{3}\left(-27-6x^{-2}\right)}{x^{7}}$
35

Split the fraction $\frac{\frac{1}{9}x^{3}\left(-27-6x^{-2}\right)+4x^{4}}{x^{7}}$ in two terms with same denominator

$0=\frac{4x^{4}}{x^{7}}+\frac{\frac{1}{9}x^{3}\left(-27-6x^{-2}\right)}{x^{7}}$
36

Simplifying the fraction by $x$

$0=4x^{\left(7\left(-1\right)+4\right)}+\frac{1}{9}x^{\left(3-7\right)}\left(-27-6x^{-2}\right)$
37

Subtract the values $3$ and $-7$

$0=4x^{\left(7\left(-1\right)+4\right)}+\frac{1}{9}x^{-4}\left(-27-6x^{-2}\right)$
38

Multiply $-1$ times $7$

$0=4x^{\left(4-7\right)}+\frac{1}{9}x^{-4}\left(-27-6x^{-2}\right)$
39

Subtract the values $4$ and $-7$

$0=4x^{-3}+\frac{1}{9}x^{-4}\left(-27-6x^{-2}\right)$
40

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

$0=4\frac{1}{x^{3}}+\frac{1}{9}\cdot\frac{1}{x^{4}}\left(-27-6\left(\frac{1}{x^{2}}\right)\right)$
41

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=-6$ and $x=x^{2}$

$0=\frac{4}{x^{3}}+\left(\frac{-6}{x^{2}}-27\right)\frac{\frac{1}{9}}{x^{4}}$

Answer

$0=\frac{4}{x^{3}}+\left(\frac{-6}{x^{2}}-27\right)\frac{\frac{1}{9}}{x^{4}}$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.47 seconds

Views:

110