We use chain rule here. Let g(x)=cosx/(1+sinx)g(x)=cosx1+sinx and f(x)=tan^(-1)xf(x)=tan−1x
then we can write h(x)=tan^(-1)(cosx/(1+sinx))h(x)=tan−1(cosx1+sinx)
as h(x)=f(g(x))h(x)=f(g(x))
and d/(dx)tan^(-1)(cosx/(1+sinx))ddxtan−1(cosx1+sinx)
= d/(dx)f(g(x))ddxf(g(x))
= (df)/(dg)xx(dg)/(dx)dfdg×dgdx
Now as f(x)=tan^(-1)xf(x)=tan−1x, (df)/(dx)=1/(1+x^2)dfdx=11+x2
and as g(x)=cosx/(1+sinx)g(x)=cosx1+sinx, using quotient rule
(dg)/(dx)=(-sinx(1+sinx)-cosx*cosx)/(1+sinx)^2dgdx=−sinx(1+sinx)−cosx⋅cosx(1+sinx)2
= (-sinx-sin^2x-cos^2x)/(1+sinx)^2=-1/(1+sinx)−sinx−sin2x−cos2x(1+sinx)2=−11+sinx
Hence d/(dx)tan^(-1)(cosx/(1+sinx))ddxtan−1(cosx1+sinx)
= 1/(1+(g(x))^2)xx(-1/(1+sinx))11+(g(x))2×(−11+sinx)
= -1/(1+(cosx/(1+sinx))^2)xx1/(1+sinx)−11+(cosx1+sinx)2×11+sinx
= -(1+sinx)/((1+sinx)^2+cos^2x)−1+sinx(1+sinx)2+cos2x
= -(1+sinx)/(1+sin^2x+2sinx+cos^2x)−1+sinx1+sin2x+2sinx+cos2x
= -1/2−12
Note: In case you are wondering why it turns out to be so simple answer, see the following.
cosx/(1+sinx)=(cos^2(x/2)-sin^2(x/2))/(cos(x/2)+sin(x/2))^2cosx1+sinx=cos2(x2)−sin2(x2)(cos(x2)+sin(x2))2
= (cos(x/2)-sin(x/2))/(cos(x/2)+sin(x/2))cos(x2)−sin(x2)cos(x2)+sin(x2)
= (1-tan(x/2))/(1+tan(x/2))=(tan(pi/4)-tan(x/2))/(1+tan(pi/4)tan(x/2))=tan(pi/4-x/2)1−tan(x2)1+tan(x2)=tan(π4)−tan(x2)1+tan(π4)tan(x2)=tan(π4−x2)
hence tan^(-1)(cosx/(1+sinx))=pi/4-x/2tan−1(cosx1+sinx)=π4−x2 and its derivative is -1/2−12
