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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 1 lim (x^{3} + 2 x^{2} - x - 2)

\int_{\frac{9}{8}}^{6} \frac{1}{1 - x ^{2}} d x

\frac{\partial}{\partial x} (\frac{( x - y )}{x y + 3})

\int cos^{4} (8 x) d x

x \to 0 lim (\frac{3 x ^{2} + 5 x}{x})

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