What is the (sqrt(x^2-1))^' ?

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

What is the first (sqrt(x^2-1))^' ?

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

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(x^{2} - 1)^{^{'}}

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

Solution steps

(x^{2} - 1)^{'}

Apply the chain rule:

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

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

(x^{2} - 1)^{'} = 2 x

= \frac{1}{2 x ^{2} - 1} \cdot 2 x

Simplify

\frac{1}{2 x ^{2} - 1} \cdot 2 x : \frac{x}{x ^{2} - 1}

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

\frac{d}{d x} f (x) = - 9

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

(cos^{4} (x))^{^{'}}

x \to 0 lim (\frac{sin ( - 4 x )}{x})

\int \frac{x ^{6}}{x ^{14} - 9} d x

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