# How do you find the derivative of arcsin((2x)/(1+x^2))?

Jul 13, 2017

d/(dx)[arcsin((2x)/(1+x^2))] = color(blue)((2(1-x^2))/((x^2+1)(sqrt(1-(4x^2)/((x^2+1)^2)))

#### Explanation:

We're asked to find the derivative

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

To do this, we can first use the chain rule, where

$\frac{d}{\mathrm{dx}} \left[\arcsin \left(\frac{2 x}{{x}^{2} + 1}\right)\right] = \frac{d}{\mathrm{du}} \left[\arcsin u\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = \frac{2 x}{{x}^{2} + 1}$

• $\frac{d}{\mathrm{du}} \left[\arcsin u\right] = \frac{1}{\sqrt{1 - {u}^{2}}}$:

$= \frac{\frac{d}{\mathrm{dx}} \left[\frac{2 x}{{x}^{2} + 1}\right]}{\sqrt{1 - \frac{4 {x}^{2}}{{\left({x}^{2} + 1\right)}^{2}}}}$

$= \frac{2 \frac{d}{\mathrm{dx}} \left[\frac{x}{1 + {x}^{2}}\right]}{\sqrt{1 - \frac{4 {x}^{2}}{{\left({x}^{2} + 1\right)}^{2}}}}$

We'll now use the quotient rule to differentiate the first quantity:

$\frac{d}{\mathrm{dx}} \left[\frac{u}{v}\right] = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{{v}^{2}}$

where

• $u = x$

• $v = {x}^{2} + 1$

= (2((x^2+1)(d/(dx)[x]) - (x)(d/(dx)[x^2+1])))/((x^2+1)(sqrt(1-(4x^2)/((x^2+1)^2)))

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

$\uparrow \uparrow$ (using the power rule, the derivative of $x$ is $1$, and the derivative of ${x}^{2} + 1$ is $2 x$)

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

= color(blue)((2(1-x^2))/((x^2+1)(sqrt(1-(4x^2)/((x^2+1)^2)))

Jul 14, 2017

dy/dx=2/(1+x^2); if |x| < 1, and,

dy/dx=-2/(1+x^2); if |x| > 1.

#### Explanation:

Let, $y = a r c \sin \left(\frac{2 x}{1 + {x}^{2}}\right) = a r c \sin u , w h e r e , u = \frac{2 x}{1 + {x}^{2}} .$

Thus, $y$ is a fun. of $u ,$ and $u$ of $x .$

Therefore, by the Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \ldots . . \left(\ast\right) .$

Now, $y = a r c \sin u \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {u}^{2}}} .$

$= \frac{1}{\sqrt{1 - {\left(\frac{2 x}{1 + {x}^{2}}\right)}^{2}}} \ldots \ldots \left[\because , u = \frac{2 x}{1 + {x}^{2}}\right] ,$

$= \frac{1}{\sqrt{\frac{{\left(1 + {x}^{2}\right)}^{2} - 4 {x}^{2}}{1 + {x}^{2}} ^ 2}} ,$

$= \frac{1 + {x}^{2}}{\sqrt{\left(1 + 2 {x}^{2} + {x}^{4}\right) - 4 {x}^{2}}} ,$

$= \frac{1 + {x}^{2}}{\sqrt{1 - 2 {x}^{2} + {x}^{4}}} = \frac{1 + {x}^{2}}{\sqrt{{\left(1 - {x}^{2}\right)}^{2}}} .$

Note that, here, the Square Root is, in fact, the

the Positive Square Root.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1 + {x}^{2}}{|} 1 - {x}^{2} | \ldots \ldots \ldots \ldots \left({\ast}^{1}\right) .$

Next, $u = \frac{2 x}{1 + {x}^{2}} \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 2 \frac{d}{\mathrm{dx}} \left\{\frac{x}{1 + {x}^{2}}\right\} ,$

$= 2 \frac{\left\{\left(1 + {x}^{2}\right) \frac{d}{\mathrm{dx}} \left(x\right) - x \frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right)\right\}}{1 + {x}^{2}} ^ 2 , \ldots \text{[Quotient Rule},}$

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

$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = \frac{2 \left(1 - {x}^{2}\right)}{1 + {x}^{2}} ^ 2. \ldots \ldots \ldots \ldots \ldots \left({\ast}^{2}\right) .$

Utilising (ast^1) & (ast^2) in $\left(\ast\right) ,$ we get,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left\{\frac{1 + {x}^{2}}{|} 1 - {x}^{2} |\right\} \left\{\frac{2 \left(1 - {x}^{2}\right)}{1 + {x}^{2}} ^ 2\right\} , i . e . ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{2}{1 + {x}^{2}}\right) \cdot \frac{1 - {x}^{2}}{|} 1 - {x}^{2} | .$

Now, recall that, $\mathmr{if} , 1 < {x}^{2} , i . e . , \left(1 - {x}^{2}\right) < 0 , | 1 - {x}^{2} | = - \left(1 - {x}^{2}\right) .$

 :. dy/dx=-2/(1+x^2); if 1 lt x^2.

Similarly, $\mathmr{if} 1 > {x}^{2} , t h e n , \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{1 + {x}^{2}} .$

Since, $1 < {x}^{2} \iff | x | < 1 ,$ we finally have,

dy/dx=2/(1+x^2); if |x| < 1, and,

dy/dx=-2/(1+x^2); if |x| > 1.

Enjoy Maths.!