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

2 Answers
May 15, 2018

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

Explanation:

show below

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

May 15, 2018

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

Explanation:

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).