How do you differentiate $f(x) =[(sin x + tan x)/(sin cos x)]^3$ ?

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Answer 1 · 2017-05-28T15:15:00

$(df)/(dx)=3((sinx+tanx)/(sinxcosx))^2(sinx(cosx+2))/cos^3x$

To differentiate $f(x)=[(sinx+tanx)/(sinxcosx)]^3$ ,

Let $f(x)=(g(x))^3$ , where $g(x)=(sinx+tanx)/(sinxcosx)$

Now as $f(x)=(g(x))^3$ , $(df)/(dg)=3(g(x))^2$

and usiing quotient rule $(dg)/(dx)=(sinxcosx xx (cosx+sec^2x)- (sinx+tanx)(cos^2x-sin^2x))/(sin^2xcos^2x)$

= $(sinxcos^2x+sinxsecx-sinxcos^2x-tanxcos^2x+sin^3x+sin^2xtanx)/(sin^2xcos^2x)$

= $(sinxcos^2x+tanx-sinxcos^2x-sinxcosx+sin^3x+sin^2xtanx)/(sin^2xcos^2x)$

= $1/sinx+1/(sinxcos^3x)-1/sinx-1/(sinxcosx)+sinx/cos^2x+sinx/cos^3x$

= $1/(sinxcos^3x)-1/(sinxcosx)+sinx/cos^2x+sinx/cos^3x$

= $(1-cos^2x)/(sinxcos^3x)+(sinx(cosx+1))/cos^3x$

= $sinx/cos^3x+(sinx(cosx+1))/cos^3x$

= $(sinx(cosx+2))/cos^3x$

and $(df)/(dx)=(df)/(dg)xx(dg)/(dx)$

= $3((sinx+tanx)/(sinxcosx))^2(sinx(cosx+2))/cos^3x$

How do you differentiate f(x) =[(sin x + tan x)/(sin cos x)]^3 f(x) =[(sin x + tan x)/(sin cos x)]^3 ?