Derive the function arcsin((1-1x^2)/(1+x^2)) with respect to x

\frac{d}{dx}\left(arcsin\left(\frac{1-x^2}{1+x^2}\right)\right)

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Answer

$\frac{-2x\left(x^2+1\right)-2x\left(1-x^2\right)}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$

Step by step solution

Problem

$\frac{d}{dx}\left(arcsin\left(\frac{1-x^2}{1+x^2}\right)\right)$
1

Taking the derivative of arcsine

$\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}\cdot\frac{d}{dx}\left(\frac{1-x^2}{x^2+1}\right)$
2

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}\cdot\frac{\left(x^2+1\right)\frac{d}{dx}\left(1-x^2\right)-\left(1-x^2\right)\frac{d}{dx}\left(x^2+1\right)}{\left(x^2+1\right)^2}$
3

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}\cdot\frac{\left(x^2+1\right)\left(\frac{d}{dx}\left(-x^2\right)+\frac{d}{dx}\left(1\right)\right)-\left(1-x^2\right)\left(\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(1\right)\right)}{\left(x^2+1\right)^2}$
4

The derivative of the constant function is equal to zero

$\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}\cdot\frac{\left(x^2+1\right)\left(\frac{d}{dx}\left(-x^2\right)+0\right)-\left(1-x^2\right)\left(\frac{d}{dx}\left(x^2\right)+0\right)}{\left(x^2+1\right)^2}$
5

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}\cdot\frac{\left(x^2+1\right)\left(0-\frac{d}{dx}\left(x^2\right)\right)-\left(1-x^2\right)\left(\frac{d}{dx}\left(x^2\right)+0\right)}{\left(x^2+1\right)^2}$
6

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{\left(x^2+1\right)\left(0-1\cdot 2x\right)-\left(2x^{\left(2-1\right)}+0\right)\left(1-x^2\right)}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}$
7

Subtract the values $2$ and $-1$

$\frac{\left(x^2+1\right)\left(0-1\cdot 2x\right)-\left(2x^{1}+0\right)\left(1-x^2\right)}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}$
8

Multiply $2$ times $-1$

$\frac{\left(x^2+1\right)\left(0-2x\right)-\left(2x^{1}+0\right)\left(1-x^2\right)}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}$
9

$x+0=x$, where $x$ is any expression

$\frac{-2x\left(x^2+1\right)-1\cdot 2\left(1-x^2\right)x^{1}}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}$
10

Multiply $2$ times $-1$

$\frac{-2x\left(x^2+1\right)-2\left(1-x^2\right)x^{1}}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}$
11

Any expression to the power of $1$ is equal to that same expression

$\frac{-2x\left(x^2+1\right)-2x\left(1-x^2\right)}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\left(\frac{1-x^2}{x^2+1}\right)^2}}$
12

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{-2x\left(x^2+1\right)-2x\left(1-x^2\right)}{\left(x^2+1\right)^2}\cdot\frac{1}{\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$
13

Multiplying fractions

$\frac{-2x\left(x^2+1\right)-2x\left(1-x^2\right)}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$

Answer

$\frac{-2x\left(x^2+1\right)-2x\left(1-x^2\right)}{\left(x^2+1\right)^2\sqrt{1-\frac{\left(1-x^2\right)^2}{\left(x^2+1\right)^2}}}$

Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.38 seconds

Views:

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