What is the (sqrt(x^2-2x+3))^' ?

The (sqrt(x^2-2x+3))^' is (x-1)/(sqrt(x^2-2x+3))

What is the first (sqrt(x^2-2x+3))^' ?

The first (sqrt(x^2-2x+3))^' is (x-1)/(sqrt(x^2-2x+3))

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(x^{2} - 2 x + 3)^{^{'}}

\frac{x - 1}{x ^{2} - 2 x + 3}

Solution steps

(x^{2} - 2 x + 3)^{'}

Apply the chain rule:

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

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

(x^{2} - 2 x + 3)^{'} = 2 x - 2

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

Simplify

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

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

\int_{- 1}^{0} x^{3} d x

- \frac{32}{x ^{3}} + \frac{4}{x ^{2}}

(1 + y e^{x y}) d x + (4 + x e^{x y}) d y = 0

\frac{d}{d x} ((arctan (2 x))^{10})

\int_{6}^{7} t t - 6 d t

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