Y=cosx+sinx/cosx-sinx find dy/dx?

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Answer 1 · 2018-05-15T17:50:15

#y'=(2(sin(x)^2+cos(x)^2))/(sin(x)-cos(x))^2#

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#y=[cosx+sinx]/[cosx-sinx]#

#y'=[(cosx-sinx)(-sinx+cosx)-(cosx+sinx)(-sinx-cosx)]/[cosx-sinx]^2#

#y'=((cos(x)-sin(x))^2-(-sin(x)-cos(x))*(sin(x)+cos(x)))/(cos(x)-sin(x))^2#

#y'=1-((-sin(x)-cos(x))*(sin(x)+cos(x)))/(cos(x)-sin(x))^2#

#y'=(2(sin^2(x)+cos^2(x)))/(sin(x)-cos(x))^2#

Answer 2 · 2018-05-15T18:01:21

#dy/dx=sec^2(pi/4+x)=2/(cosx-sinx)^2=2/(1-sin2x)#.

I presume that, #y=(cosx+sinx)/(cosx-sinx)#,

#={cosx(1+sinx/cosx)}/{cosx(1-sinx/cosx)}#,

#=(1+tanx)/(1-tanx)#,

# rArr y=tan(pi/4+x)#

#:. dy/dx=sec^2(pi/4+x)*d/dx(pi/4+x)..."[The Chain Rule]"#,

#=sec^2(pi/4+x)#.

Also, #dy/dx=1/cos^2(pi/4+x)#,

#=1/(cos(pi/4+x))^2#,

#=1/(cos(pi/4)cosx-sin(pi/4)sinx)^2#,

#=1/(1/sqrt2*cosx-1/sqrt2*sinx)^2#.

# rArr dy/dx=2/(cosx-sinx)^2, or, #

#dy/dx=2/(cos^2x-2cosxsinx+sin^2x)=2/(1-sin2x)#.