trig. differentiation question (1 Viewer)

[freaking_out]

differentiate the following:

(cosx+sinx)/(cosx-sinx)

and show that the derivative is equal to sec^2(pi/4+x)


thanx, in advance

 

[Constip8edSkunk]

LHS=d/dx[(cosx+sinx)/(cosx-sinx)]
=[(cosx-sinx)^2+(cosx+sinx)^2]/(cosx-sinx)^2 --using quotient rule
=2(cos^2x+sin^2x)/1-2cosxsinx
=2/(1-2cosxsinx)

RHS=sec^2(pi/4+x)
=1/cos^2(pi/4+x)
=1/[cos(pi/4)cosx-sin(pi/4)sinx]^2
=1/[cos^2(pi/4)(cos^2x+sin^2x-2cosxsinx)] --sin(pi/4)=cos(pi/4)
=1/[1/2(1-2cosxsinx)]
=2/(1-2cosxsinx)
=LHS

LHS=RHS, therefor d/dx[(cosx+sinx)/(cosx-sinx)] = sec^2(pi/4+x)

the algorithms pretty cumbersome so theres prolly a faster, better way, hmmm......