# Derive the function (3x^2)/(2y) with respect to x

## \frac{d}{dx}\left(\frac{3x^2}{2y}\right)

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$\frac{9x}{y^{2}}$

## Step by step solution

Problem

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

Taking out the constant $2$ from the fraction's denominator

$\frac{d}{dx}\left(\frac{\frac{1}{2}\cdot\frac{3x^2}{y}}{2}\right)$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

The derivative of the constant function is equal to zero

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

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

$\frac{0+2\frac{d}{dx}\left(\frac{1}{2}\cdot\frac{3x^2}{y}\right)}{4}$
5

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

$\frac{0+2\cdot \frac{1}{2}\cdot\frac{d}{dx}\left(\frac{3x^2}{y}\right)}{4}$
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{0+2\cdot \frac{1}{2}\left(\frac{y\frac{d}{dx}\left(3x^2\right)-3x^2\frac{d}{dx}\left(y\right)}{y^2}\right)}{4}$
7

The derivative of the constant function is equal to zero

$\frac{0+2\cdot \frac{1}{2}\left(\frac{0\left(-3\right)x^2+y\frac{d}{dx}\left(3x^2\right)}{y^2}\right)}{4}$
8

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

$\frac{0+2\cdot \frac{1}{2}\left(\frac{0+y\frac{d}{dx}\left(3x^2\right)}{y^2}\right)}{4}$
9

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

$\frac{0+2\cdot \frac{1}{2}\left(\frac{0+3y\frac{d}{dx}\left(x^2\right)}{y^2}\right)}{4}$
10

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

$\frac{0+2\cdot \frac{1}{2}\left(\frac{0+3\cdot 2y\cdot x}{y^2}\right)}{4}$
11

Multiply $2$ times $3$

$\frac{0+1\frac{0+6y\cdot x}{y^2}}{4}$
12

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

$\frac{\frac{6y\cdot x}{y^2}}{4}$
13

Simplifying the fraction by $y$

$\frac{\frac{6x}{y}}{4}$
14

Simplifying the fraction

$\frac{6x}{4y}$
15

Apply the formula: $\frac{b\cdot a}{f\cdot c}$$=\frac{b}{f}\cdot\frac{a}{c}, where a=6, b=x, c=4 and f=y \frac{6\cdot \frac{3}{2}\left(\frac{x}{y}\right)}{1y} 16 Multiply \frac{3}{2} times 6 \frac{9\frac{x}{y}}{1y} 17 Any expression multiplied by 1 is equal to itself \frac{9\frac{x}{y}}{y} 18 Multiplying the fraction and term \frac{\frac{9x}{y}}{y} 19 Simplifying the fraction \frac{9x}{y\cdot y} 20 When multiplying exponents with same base you can add the exponents \frac{9x}{y^2} 21 Rewrite the exponent using the power rule \frac{a^m}{a^n}=a^{m-n}, where in this case m=0 9xy^{-2} 22 Applying the property of exponents, \displaystyle a^{-n}=\frac{1}{a^n}, where n is a number 9x\frac{1}{y^{2}} 23 Apply the formula: a\frac{1}{x}$$=\frac{a}{x}$, where $a=9$ and $x=y^{2}$

$x\frac{9}{y^{2}}$
24

Multiplying the fraction and term

$\frac{9x}{y^{2}}$

$\frac{9x}{y^{2}}$

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### Main topic:

Differential calculus

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