Find the derivative of (x^2)/(2x-1)

\frac{d}{dx}\left(\frac{x^2}{2x-1}\right)

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Answer

$\frac{2x\left(2x-1\right)-2x^2}{\left(2x-1\right)^2}$

Step by step solution

Problem

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

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

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

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

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

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

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

The derivative of the constant function is equal to zero

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

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

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

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

$\frac{\left(0+1\cdot 2\right)\left(-1\right)x^2+2x\left(2x-1\right)}{\left(2x-1\right)^2}$
7

Multiply $2$ times $1$

$\frac{\left(0+2\right)\left(-1\right)x^2+2x\left(2x-1\right)}{\left(2x-1\right)^2}$
8

Add the values $2$ and $0$

$\frac{2\left(-1\right)x^2+2x\left(2x-1\right)}{\left(2x-1\right)^2}$
9

Multiply $-1$ times $2$

$\frac{2x\left(2x-1\right)-2x^2}{\left(2x-1\right)^2}$

Answer

$\frac{2x\left(2x-1\right)-2x^2}{\left(2x-1\right)^2}$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.26 seconds

Views:

66