Question

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

Answer 1

The given function is f(x) = $\frac {sin\ (ax+b)}{cos\ (cx+d)}$ = $\frac {g(x)}{h(x)}$, where g(x) = sin (ax+b) and h(x) = cos (cx+d)
∴f' = $\frac {g'h-gh'}{h_2}$
consider g(x) = sin (ax+b)
Let u(x) = ax+b, v(t) = sin t
Then, (vou)(x) = v(u(x)) = v(ax+b) = sin (ax+b) = g(x)
∴ g is a composite function of two functions, u and v

Put t = u(x) = ax+b
$\frac {dv}{dt}$ = $\frac {d}{dt}$ (sin t) = cost = cos (ax+b)
$\frac {dt}{dx}$ = $\frac {d}{dx}$(ax+b) = $\frac {d}{dx}$(ax) + $\frac {d}{dx}$(b) = a+0 = a
Therefore, by chain rule, we obtain
g'=$\frac {dg}{dx}$ = $\frac {dv}{dt}$.$\frac {dt}{dx}$ = cos (ax+b).a = a cos (ax+b)
Consider h(x) = cos (cx+d)
Let p(x) = cx+d, q(y) = cos y
Then, (qop)(x) = q(p(x)) = q(cx+d) = cos (cx+d) = h(x)

∴h is a composite function of two functions, p and q

Put y = p(x) = cx+d
$\frac {dq}{dy}$=$\frac {d}{dy}$ (cos y) = -siny = -sin (cx+d)
$\frac {dy}{dx}$=$\frac {d}{dx}$(cx+d) =$\frac {d}{dx}$(cx) + $\frac {d}{dx}$(d) = c
Therefore, by chain rule, we obtain
h'=$\frac {dh}{dx}$=$\frac {dq}{dy}$.$\frac {dy}{dx}$ = -sin (cx+d) . c = -c sin (cx+d)
∴f'= $\frac {acos \ (ax+b) . cos (cx+d) - sin\ (ax+b){-c sin\ (cx+d)}}{[cos\ (cx+d)]^2}$
=$\frac {acos\ (ax+b)}{cos\ (cx+d)}$ + csin (ax+b).$\frac {sin\ (cx+d)}{cos\ (cx+d)}$.$\frac {1}{cos\ (cx+d)}$
=acos (ax+b).sec (cx+d) + csin (ax+b).tan (cx+d).sec (cx+d)