How to Differentiate #t (1+t^2)^(-1/2)#? Thnx

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Answer 1 · 2017-12-31T15:55:09

#y=t (1+t^2)^(-1/2)#

Let #t=tanx#

Differentiating w r to x we get

#(dt)/(dx)=sec^2x=1+tan^2x=1+t^2#

So

#y=tanx(1+tan^2x)^(-1/2)#

#=>y=sinx/cosx(sec^2x)^(-1/2)#

#=>y=sinxsecx(secx)^(-1)#

#=>y=sinx#

Differentiating w r to x we get

#(dy)/(dx)=cosx=1/secx#

#=1/(1+tan^2x)^(1/2)#

#=1/(1+t^2)^(1/2)=(1+t^2)^(-1/2)#

Now

#(d y)/(dt)=((dy)/(dx))/((dt)/(dx))=((1+t^2)^(-1/2))/(1+t^2)= (1+t^2)^(-3/2) #