Question

Differentiate the following functions with respect to x

$\frac{sin(ax+b)}{cos(cx+d)}$

Answer 1

Ley y = $\frac{sin(ax+b)}{cos(cx+d)}$

Differentiate both sides w.r.t. x, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{sin(ax+b)}{cos(cx+d)}\right)$

= $\frac{cos(cx+d)\frac{d}{dx}\left\{sin\right(ax+b\left)\right\}-sin(ax+b)\frac{d}{dx}\left\{cos\right(cx+d\left)\right\}}{\left\{cos\right(cx+d){\}}^2}$

= $\frac{cos(cx+d)cos(ax+b).(a+0)+sin(ax+b)sin(cx+d)(c+0)}{co{s}^2(cx+d)}$

$\left[\begin{array}{c}Chainrule-\\ \frac{d}{dx}sin(ax+b)=cos(ax+b)\frac{d}{dx}(ax+b)\\ =cos(ax+b)\times (a\times 1+0)\\ \frac{d}{dx}cos(cx+d)=-sin(cx+d)\frac{d}{dx}(cx+d)\\ =-sin(cx+d)\times (c\times 1+0)\end{array}\right]$

=$\frac{acos(cx+d)cos(ax+b)+csin(ax+b)sin(cx+d)}{co{s}^2(cx+d)}$

= $\frac{acos(cx+d)cos(ax+b)}{co{s}^2(cx+d)}+\frac{csin(ax+b)sin(cx+d)}{co{s}^2(cx+d)}$

= $\frac{acos(ax+b)}{cos(cx+d)}+\frac{csin(ax+b)sin(cx+d)}{cos(cx+d)cos(cx+d)}$

=$acos(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d)$