What is the derivative of #(1/(ab)) tan^-1((b/a)tanx)#?

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Answer 1 · 2017-04-25T05:51:37

#" The Reqd. Deri.="1/(a^2cos^2x+b^2sin^2x).#

Let, #y=1/(ab)tan^-1{(b/a)tanx}.#

We will use the Chain Rule and Standard Derivative of #tan^-1x#.

Therefore, #dy/dx=1/(ab)d/dx[tan^-1{(b/a)tanx}],#

#=1/(ab){1/{1+((b/a)tanx)^2}}[d/dx{(b/a)tanx},#

#=1/(ab){a^2/(a^2+b^2tan^2x)}{(b/a)sec^2x},#

#=sec^2x/(a^2+b^2tan^2x)#

#=(1/cos^2x)/{a^2+b^2(sin^2x/cos^2x)},#

#:." The Reqd. Deri.="1/(a^2cos^2x+b^2sin^2x).#

Enjoy Maths.!

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