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

$\begin{array}{1 1} a.sin(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d) \\ a.sin(ax+b)sec(cx+d)-csin(ax+b)tan(cx+d)sec(cx+d) \\ a.sin(ax+b)cosec(cx+d)+csin(ax+b)cot(cx+d)sec(cx+d) \\ a.sin(ax+b)cosec(cx+d)-csin(ax+b)cot(cx+d)sec(cx+d)\end{array}$

## 1 Answer

Toolbox:
• Whenever we see a function of the form $y = \large \frac{u}{v}$, which is a quotient of two other functions with derivatives, we can apply the following quotient rule: $y'=\large \frac{1}{v^2}\;$$( v\large\frac{du}{dx}$$-u \large \frac{dv}{dx}$$) • \; \large \frac{d(sinx)}{dx}$$= cosx$
• $\; \large \frac{d(cosx)}{dx} $$=-sinx • According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). Given y =\large \frac{ sin\; (ax + b) } { cos \;( cx + d)}, let us take u = sin\; (ax + b) and v = cos \;( cx + d) Whenever we see a function of the form y = \large \frac{u}{v}, which is a quotient of two other functions with derivatives, we can apply the following quotient rule: y'=\large \frac{1}{v^2}\;$$ ( v\large\frac{du}{dx}$$-u \large \frac{dv}{dx}$$)$
$\textbf{Step 1}$:
$\Rightarrow v\large\frac{du}{dx} $$= cos(cx+d). d(sinx(ax+b)) According to the Chain Rule for differentiation, given two functions f(x) and g(x), and y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x). \; \large \frac{d(sinx)}{dx}$$= cosx$
We need to differentiate $sin(ax+b)$ using the chain rule.
In this case, Let $g(x) = ax+b \rightarrow g'(x) = a$
$\Rightarrow f'(g(x)) = f'(sinx(ax+b)) = cos(ax+b)$
Therefore $d(sin(ax+b) = a. cos(ax+b)$
$\Rightarrow v\large\frac{du}{dx} $$= a. cos(cx+d).sin(ax+b) \textbf{Step 2}: \Rightarrow u \large \frac{dv}{dx}$$ = sin(ax+b). d(cos(cx+d))$
According to the Chain Rule for differentiation, given two functions $f(x)$ and $g(x)$, and $y=f(g(x)) \rightarrow y' = f'(g(x)).g'(x)$.
$\; \large \frac{d(cosx)}{dx} $$=-sinx We need to differentiate cos(cx+d) using the chain rule. In this case, Let h(x) = cx+d \rightarrow h'(x) = c \Rightarrow f'h(x) = f'(cx+d) = -sin(cx+d) Therefore d(cos(cx+d) = -c.sin(cx+d) \Rightarrow u \large \frac{dv}{dx}$$ = -c.sin(ax+b).sin(cx+d)$
$\textbf{Step 3}$:
Putting it all together, we get $y' = \large \frac{1}{(cos(cx+d))^2}$$[a.cos(cx+d).sin(ax+b)- (-c.sin(ax+b).sin(cx+d))] \Rightarrow y' = a\large \frac{sin(ax+b)}{cos(cx+d)}$$+ c \large\frac{sin(ax+b) sin(cx+d)}{cos(cx+d)cos(cx+d)}$
$\Rightarrow y' = a.sin(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d)$
answered Apr 4, 2013

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