Dy/dx of tan inverse of (1+sinx)/(1-sinx) ?

1 Answer
Steve M
Mar 28, 2018

# d/dx arctan( (1+sinx)/(1-sinx) ) = ( cosx ) / (1+sin^2x)#

Explanation:

We seek:

# d/dx arctan( (1+sinx)/(1-sinx) ) #

For simplicity, define #y=arctan( (1+sinx)/(1-sinx) )#, and We use the standard result:

# d/dx arctanx=1/(1+x^2)#

in conjunction with the chain rule (and quotient rule) to get:

# dy/dx = 1/(1+((1+sinx)/(1-sinx))^2) * d/dx ( (1+sinx)/(1-sinx) )#

# \ \ \ \ \ = 1/(1+(1+sinx)^2/(1-sinx)^2) * { ( (1-sinx)d/dx(1+sinx) - d/dx(1-sinx)(1+sinx) ) / (1-sinx)^2}#

# \ \ \ \ \ = (1/((1-sinx)^2+(1+sinx)^2))/(1-sinx)^2 (( (1-sinx)cosx + cosx(1+sinx) ) / (1-sinx)^2 )#

# \ \ \ \ \ = ((1-sinx)^2/((1-sinx)^2+(1+sinx)^2)) (( (1-sinx)cosx + cosx(1+sinx) ) / (1-sinx)^2)#

# \ \ \ \ \ = ( (1-sinx)cosx + cosx(1+sinx) ) / ((1-sinx)^2+(1+sinx)^2)#

# \ \ \ \ \ = ( cosx-sinxcosx + cosx+sinxcosx ) / (1-2sinx+sin^2x+1+2sinx+sin^2x)#

# \ \ \ \ \ = ( 2cosx ) / (2+2sin^2x)#

# \ \ \ \ \ = ( cosx ) / (1+sin^2x)#

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