5. Differentiate the functions with respect to x in 

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

Answers (1)

Given function is
f(x) = \frac{\sin (ax +b )}{\cos (cx + d)} = \frac{g(x)}{h(x)}
We know that,
f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)}
g(x) = \sin(ax+b)    and      h(x) = \cos(cx+d)
Lets take  u = (ax+b) \ and \ v = (cx+d)
Then,  
\sin (ax+b) = \sin u \ and \ \cos(cx+d) = \cos c
g^{'}(x)=\frac{d(g(x))}{dx} = \frac{d(g(x))}{du}.\frac{du}{dx}               (By chain rule)
\frac{d(g(x))}{du} = \frac{d(\sin u)}{du} = \cos u = \cos(ax+b) \ \ \ \ \ \ \ \ \ (\because u = ax +b)
\frac{du}{dx} = \frac{d(ax+b)}{dx} = a
 g^{'}(x)=a\cos (ax+b)                         -(i)
Similarly, 
h^{'}(x)=\frac{d(h(x))}{dx} = \frac{d(h(x))}{dv}.\frac{dv}{dx}
\frac{d(h(x))}{dv}= \frac{d(\cos v)}{dv} = -\sin v = -\sin (cx+d) \ \ \ \ \ \ \ (\because v = (cx+d))
\frac{dv}{dx}= \frac{d(cx+d)}{dv} = c
h^{'}(x)=-c\sin(cx+d)                      -(ii)
Now, put (i) and (ii) in  
f^{'}(x) = \frac{g^{'}(x)h(x)-g(x)h^{'}(x)}{h^2(x)} = \frac{a\cos(ax+b).\cos(cx+d)-\sin(ax+b).(-c.\sin(cx+d))}{\cos^2(cx+d)}
                                                                   = \frac{a\cos(ax+b).\cos(cx+d)}{\cos^2(cx+d)}+\frac{\sin(ax+b).c.\sin(cx+d)}{\cos^2(cx+d)}
                                                                   = a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)
Therefore, the answer is a\cos(ax+b).\sec(cx+d) +c\sin(ax+b).\tan(cx+d).\sec(cx+d)

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