Question

Differentiate the function with respect to $x$
$\frac{\sin (ax+b)}{\cos (cx+d)}$

Answer 1

We have, $f\left(x\right)=\frac{\sin (ax+b)}{\cos (cx+d)}$
Thus using quotient rule and chain rule simualtaneously,
$f^{\prime }\left(x\right)=\frac{a\cos (ax+b).\cos (cx+d)-\sin (ax+b)(-c\sin (cx+d\left)\right)}{\left[\cos \right(cx+d){]}^2}$
$=\frac{a\cos (ax+b)}{\cos (cx+d)}+c\sin (ax+b).\frac{\sin (cx+d)}{\cos (cx+d)}\times \frac{1}{\cos (cx+d)}$
$=a\cos (ax+b)\sec (cx+d)+c\sin (ax+b)\tan (cx+d)\sec (cx+d)$