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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{df}{d x} f (x) = cos (2 x)

(cos (7 x)) y^{^{'}} - 7 (tan (7 x)) y = 9 sec (7 x)

\int_{1}^{2} \frac{3}{x ^{4}} d x

\frac{d r}{d θ} = \frac{r ^{2} + 4 r θ}{2 θ ^{2}}

d er i v a t i v e sin (3 x^{2} + 2)

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