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}} (- 2 x)

Simplify

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

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

y^{^{'}} + x y + 4 = 0

\int \frac{1}{x ^{2} - 36} d x

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

\frac{d}{d x} (f (h, x) (g (h, x) (h (x))))

\int x^{2} + 1 - \frac{1}{2} + \frac{1}{16 x ^{2}} d 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))