Question

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

Answer 1

`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

`d/(dx) [arcsin((2x)/(1+x^2))]`

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

`d/(dx) [arcsin((2x)/(x^2 + 1))] = d/(du)[arcsinu] (du)/(dx)`

where

• `u = (2x)/(x^2 + 1)`

• `d/(du) [arcsinu] = 1/(sqrt(1-u^2))`:

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

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

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

`d/(dx) [u/v] = (v(du)/(dx) - u(dv)/(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)))`

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

`uarruarr` (using the power rule, the derivative of `x` is `1`, and the derivative of `x^2 + 1` is `2x`)

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

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

Answer 2

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

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

Explanation:

Let, `y=arc sin((2x)/(1+x^2))=arc sinu, where, u=(2x)/(1+x^2).`

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

Therefore, by the Chain Rule,

`dy/dx=dy/(du)*(du)/dx.................(ast).`

Now, `y=arc sin u rArr dy/(du)=1/sqrt(1-u^2).`

`=1/sqrt{1-((2x)/(1+x^2))^2}......[because, u=(2x)/(1+x^2)],`

`=1/sqrt{((1+x^2)^2-4x^2)/(1+x^2)^2},`

`=(1+x^2)/sqrt{(1+2x^2+x^4)-4x^2),`

`=(1+x^2)/sqrt(1-2x^2+x^4)=(1+x^2)/sqrt{(1-x^2)^2}.`

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

the Positive Square Root.

`rArr dy/(du)=(1+x^2)/|1-x^2|............(ast^1).`

Next, `u=(2x)/(1+x^2) rArr (du)/dx=2d/dx{x/(1+x^2)},`

`=2[{(1+x^2)d/dx(x)-xd/dx(1+x^2)}]/(1+x^2)^2,..."[Quotient Rule},"`

`=[2{(1+x^2)(1)-x(2x)}]/(1+x^2)^2.`

`rArr (du)/dx=(2(1-x^2))/(1+x^2)^2................(ast^2).`

Utilising `(ast^1) & (ast^2)` in `(ast),` we get,

`dy/dx={(1+x^2)/|1-x^2|}{(2(1-x^2))/(1+x^2)^2}, i.e.,`

`dy/dx=(2/(1+x^2))*(1-x^2)/|1-x^2|.`

Now, recall that, `if, 1 < x^2, i.e., (1-x^2) <0, |1-x^2|=-(1-x^2).`

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

Similarly, `if 1 gt x^2, then, dy/dx=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.!