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

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

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Answer 1 · 2017-05-28T15:15:00

$\frac{d f}{d x} = 3 {(\frac{sin x + tan x}{sin x cos x})}^{2} \frac{sin x (cos x + 2)}{{cos}^{3} x}$ $(df)/(dx)=3((sinx+tanx)/(sinxcosx))^2(sinx(cosx+2))/cos^3x$

Explanation:

To differentiate $f (x) = {[\frac{sin x + tan x}{sin x cos x}]}^{3}$ $f(x)=[(sinx+tanx)/(sinxcosx)]^3$ ,

we use chain rule and quotient rule.

Let $f (x) = {(g (x))}^{3}$ $f(x)=(g(x))^3$ , where $g (x) = \frac{sin x + tan x}{sin x cos x}$ $g(x)=(sinx+tanx)/(sinxcosx)$

Now as $f (x) = {(g (x))}^{3}$ $f(x)=(g(x))^3$ , $\frac{d f}{d g} = 3 {(g (x))}^{2}$ $(df)/(dg)=3(g(x))^2$

and usiing quotient rule $\frac{d g}{d x} = \frac{sin x cos x \times (cos x + {sec}^{2} x) - (sin x + tan x) ({cos}^{2} x - {sin}^{2} x)}{{sin}^{2} x {cos}^{2} x}$ $(dg)/(dx)=(sinxcosx xx (cosx+sec^2x)- (sinx+tanx)(cos^2x-sin^2x))/(sin^2xcos^2x)$

= $\frac{sin x {cos}^{2} x + sin x sec x - sin x {cos}^{2} x - tan x {cos}^{2} x + {sin}^{3} x + {sin}^{2} x tan x}{{sin}^{2} x {cos}^{2} x}$ $(sinxcos^2x+sinxsecx-sinxcos^2x-tanxcos^2x+sin^3x+sin^2xtanx)/(sin^2xcos^2x)$

= $\frac{sin x {cos}^{2} x + tan x - sin x {cos}^{2} x - sin x cos x + {sin}^{3} x + {sin}^{2} x tan x}{{sin}^{2} x {cos}^{2} x}$ $(sinxcos^2x+tanx-sinxcos^2x-sinxcosx+sin^3x+sin^2xtanx)/(sin^2xcos^2x)$

= $\frac{1}{sin x} + \frac{1}{sin x {cos}^{3} x} - \frac{1}{sin x} - \frac{1}{sin x cos x} + \frac{sin x}{{cos}^{2} x} + \frac{sin x}{{cos}^{3} x}$ $1/sinx+1/(sinxcos^3x)-1/sinx-1/(sinxcosx)+sinx/cos^2x+sinx/cos^3x$

= $\frac{1}{sin x {cos}^{3} x} - \frac{1}{sin x cos x} + \frac{sin x}{{cos}^{2} x} + \frac{sin x}{{cos}^{3} x}$ $1/(sinxcos^3x)-1/(sinxcosx)+sinx/cos^2x+sinx/cos^3x$

= $\frac{1 - {cos}^{2} x}{sin x {cos}^{3} x} + \frac{sin x (cos x + 1)}{{cos}^{3} x}$ $(1-cos^2x)/(sinxcos^3x)+(sinx(cosx+1))/cos^3x$

= $\frac{sin x}{{cos}^{3} x} + \frac{sin x (cos x + 1)}{{cos}^{3} x}$ $sinx/cos^3x+(sinx(cosx+1))/cos^3x$

= $\frac{sin x (cos x + 2)}{{cos}^{3} x}$ $(sinx(cosx+2))/cos^3x$

and $\frac{d f}{d x} = \frac{d f}{d g} \times \frac{d g}{d x}$ $(df)/(dx)=(df)/(dg)xx(dg)/(dx)$

= $3 {(\frac{sin x + tan x}{sin x cos x})}^{2} \frac{sin x (cos x + 2)}{{cos}^{3} x}$ $3((sinx+tanx)/(sinxcosx))^2(sinx(cosx+2))/cos^3x$

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

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Explanation:

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