What is the derivative of 1/(2x^2-3x+1) ?

The derivative of 1/(2x^2-3x+1) is -(4x-3)/((2x^2-3x+1)^2)

What is the first derivative of 1/(2x^2-3x+1) ?

The first derivative of 1/(2x^2-3x+1) is -(4x-3)/((2x^2-3x+1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{1}{2 x ^{2} - 3 x + 1})

- \frac{4 x - 3}{( 2 x ^{2} - 3 x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{2 x ^{2} - 3 x + 1})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((2 x^{2} - 3 x + 1)^{- 1})

Apply the chain rule:

- \frac{1}{( 2 x ^{2} - 3 x + 1 ) ^{2}} \frac{d}{d x} (2 x^{2} - 3 x + 1)

= - \frac{1}{( 2 x ^{2} - 3 x + 1 ) ^{2}} \frac{d}{d x} (2 x^{2} - 3 x + 1)

\frac{d}{d x} (2 x^{2} - 3 x + 1) = 4 x - 3

= - \frac{1}{( 2 x ^{2} - 3 x + 1 ) ^{2}} (4 x - 3)

Simplify

- \frac{1}{( 2 x ^{2} - 3 x + 1 ) ^{2}} (4 x - 3) : - \frac{4 x - 3}{( 2 x ^{2} - 3 x + 1 ) ^{2}}

= - \frac{4 x - 3}{( 2 x ^{2} - 3 x + 1 ) ^{2}}

n = 1 \sum \infty \frac{n ^{5}}{5 ^{n}}

\int_{x^{2}}^{4 - x^{2}} y d y

\int \frac{1}{( x ^{2} + 4 ) ^{\frac{5}{2}}} d x

y = 2 x e^{- 3 x} + 3 e^{x^{2}}

y^{^{''}} = 3 t^{2}

What is the derivative of 1/(2x^2-3x+1) ?
The derivative of 1/(2x^2-3x+1) is -(4x-3)/((2x^2-3x+1)^2)
What is the first derivative of 1/(2x^2-3x+1) ?
The first derivative of 1/(2x^2-3x+1) is -(4x-3)/((2x^2-3x+1)^2)