Question #48c60

2 Answers
Jim G.
Apr 13, 2017

#-(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#

Ratnaker Mehta
Apr 13, 2017

# -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).#

Enjot Maths.!

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
2740 views around the world
You can reuse this answer
Creative Commons License