Question
Differentiate the function with respect to x : $\frac{{\sin \left( {ax + b} \right)}}{{\cos \left( {cx + d} \right)}}$
Using quotient rule,
$\therefore \frac{{dy}}{{dx}} = \frac{{\cos \left( {cx + d} \right)\frac{d}{{dx}}\sin \left( {ax + b} \right) - \sin \left( {ax + b} \right)\frac{d}{{dx}}\cos \left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= \frac{{\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)\frac{d}{{dx}}\left( {ax + b} \right) - \sin \left( {ax + b} \right)\left\{ { - \sin \left( {cx + d} \right)} \right\}\frac{d}{{dx}}\left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= \frac{{\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)\left( a \right) + \sin \left( {ax + b} \right)\sin \left( {cx + d} \right)\left( c \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= \frac{{a.\cos \left( {cx + d} \right)\cos \left( {ax + b} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}} + \frac{{c.\sin \left( {ax + b} \right)\sin \left( {cx + d} \right)}}{{{{\cos }^2}\left( {cx + d} \right)}}$
$= a.\cos \left( {ax + b} \right).\sec \left( {cx + d} \right) + c.\sin \left( {ax + b} \right).\tan \left( {cx + d} \right).\sec \left( {cx + d} \right)$
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