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

2 Answers
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

#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)))#

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=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.!