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

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

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

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

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

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

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