Question #48c60

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Question #48c60

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Answer 1 · 2017-04-13T10:12:39

$-(2sec^2x)/(1+tanx)^2$

Explanation:

differentiate using the $color(blue)"quotient rule"$

$"Given " f(x)=(g(x))/(h(x))" then"$

$• f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2$

$color(orange)"Reminder" d/dx(tanx)=sec^2x$

$"here " g(x)=1-tanxrArrg'(x)=-sec^2x$

$"and " h(x)=1+tanxrArrh'(x)=sec^2x$

$rArrf'(x)=((1+tanx)(-sec^2x)-(1-tanx)(sec^2x))/(1+tanx)^2$

$color(white)(rArrf'(x))=(-sec^2xcancel(-sec^2xtanx)-sec^2xcancel(+sec^2xtanx))/(1+tanx)^2$

$color(white)(rArrf'(x))=-(2sec^2x)/(1+tanx)^2$

Answer 2 · 2017-04-13T10:45:56

$-2/(cosx+sinx)^2,$
or,
$-2/(1+sin2x).$

Explanation:

Observe that, $(1-tanx)/(1+tanx)=tan(pi/4-x).$

Hence, $d/dx{(1-tanx)/(1+tanx)}=d/dx{tan(pi/4-x),$

$=sec^2(pi/4-x)*d/dx(pi/4-x),...[because," The Chain Rule]"$

$=-sec^2(pi/4-x),$

$=-1/{cos(pi/4-x)}^2,$

$-1/{cos(pi/4)cosx+sin(pi/4)sinx}^2,$

$=-1/{1/sqrt2(cosx+sinx)}^2,$

$=-2/(cosx+sinx)^2, or,$

$=-2/{cos^2x+2cosxsinx+sin^2x},$

$=-2/(1+sin2x).$

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Question #48c60

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