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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

4 y^{^{'}} - 12 x^{2} = 0

\frac{\partial}{\partial x} (x sin^{2} (x y - y))

\int \frac{3}{4 x} d x

x^{2} - 4 x + 2

n = 0 \sum \infty \frac{1}{n ( n - 1 )}

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