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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = 7 x, y = x^{2}

\int \frac{5 sec ( θ ) tan ( θ )}{sec ^{2} ( θ ) - sec ( θ )} d θ

\int \frac{3 ^{t a n (5 x)}}{cos ^{2} ( 5 x )} d x

\frac{( t ^{3} + t )}{t ^{4} - 2}

x \to 2 + lim (f (x))

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