Question
The derivative of $${\text{ln}}\left( {x + \sin \,x} \right)$$ with respect to $$\left( {x + \cos \,x} \right)$$ is :
A.
$$\frac{{1 + \cos \,x}}{{\left( {x + \sin \,x} \right)\left( {1 - \sin \,x} \right)}}$$
B.
$$\frac{{1 - \cos \,x}}{{\left( {x + \sin \,x} \right)\left( {1 + \sin \,x} \right)}}$$
C.
$$\frac{{1 - \cos \,x}}{{\left( {x - \sin \,x} \right)\left( {1 + \cos \,x} \right)}}$$
D.
$$\frac{{1 + \cos \,x}}{{\left( {x - \sin \,x} \right)\left( {1 - \cos \,x} \right)}}$$
Answer :
$$\frac{{1 + \cos \,x}}{{\left( {x + \sin \,x} \right)\left( {1 - \sin \,x} \right)}}$$
Solution :
$$\eqalign{
& {\text{ln}}{\left( {x + \sin \,x} \right)^1} = y\,\,\,\,\,\,\,\,\,\left( {{\text{say}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\left( {x + \sin \,x} \right)}}\left( {1 + \cos \,x} \right) = \frac{{\left( {1 + \cos \,x} \right)}}{{\left( {x + \sin \,x} \right)}} \cr
& x + \cos \,x = z\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{say}}} \right) \cr
& \frac{{dz}}{{dx}} = \left( {1 - \sin \,x} \right) \cr
& {\text{derivative of ln}}\left( {x + \sin \,x} \right){\text{ w}}{\text{.r}}{\text{.t }}\left( {x + \cos \,x} \right){\text{ is}} \cr
& \frac{{dy}}{{dz}} = \frac{{\left( {1 + \cos \,x} \right)}}{{\left( {x + \sin \,x} \right)\left( {1 - \sin \,x} \right)}} \cr} $$