Differentiate the function with respect to \[x\]: \[\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}\]

Hint: To differentiate the functions of this type, we will use the quotient rule. According to quotient rule, \[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{\left( {g(x) \times f'(x)} \right) - \left( {f(x) \times g'(x)} \right)}}{{{{\left( {g(x)} \right)}^2}}}\], where \[f'(x) = \dfrac{d}{{dx}}\left( {f(x)} \right)\]. Also, we see that \[f(x)\] and \[g(x)\] are composite functions. We will again use chain rule to find their derivatives and then substitute in the formula for quotient rule. According to chain rule, \[\dfrac{d}{{dx}}\left( {mon(x)} \right) = m'(n(x)) \times n'(x)\], where \[m'(x) = \dfrac{d}{{dx}}\left( {m(x)} \right)\]. After finding the values of these composite functions, we will then substitute back in the formula for the quotient rule.

Complete answer: We need to differentiate \[\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}\] with respect to \[x\]. i.e. we need to find \[\dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right)\].
Let \[\sin (ax + b) = f(x) - - - - - - (1)\]
\[\cos (cx + d) = g(x) - - - - - - (2)\]
Now Putting these values in quotient rule, we have
\[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right)\]
We know, \[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{\left( {g(x) \times f'(x)} \right) - \left( {f(x) \times g'(x)} \right)}}{{{{\left( {g(x)} \right)}^2}}}\], where \[f'(x) = \dfrac{d}{{dx}}\left( {f(x)} \right)\]. So,
\[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( {\cos (cx + d) \times \dfrac{d}{{dx}}\left( {\sin (ax + b)} \right)} \right) - \left( {\sin (ax + b) \times \dfrac{d}{{dx}}\left( {\cos (cx + d)} \right)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}} - - (2)\]
We will first find \[\dfrac{d}{{dx}}\left( {\sin (ax + b)} \right)\]
We see that this function is the composition of two functions.
Let \[m(x) = \sin x - - - - - - (3)\]
And \[n(x) = ax + b - - - - - - (4)\]
Hence, \[mon(x) = m\left( {n(x)} \right) = \sin \left( {n(x)} \right) = \sin \left( {ax + b} \right)\]
Using chain rule, we have
\[\dfrac{d}{{dx}}\left( {mon(x)} \right) = m'(n(x)) \times n'(x)\], where \[m'(x) = \dfrac{d}{{dx}}\left( {m(x)} \right) - - - - - - (5)\]
Hence, \[\dfrac{d}{{dx}}\left( {mon(x)} \right) = \dfrac{d}{{dx}}\left( {\sin (ax + b)} \right) - - - - - - (6)\].
First of all, finding \[m'(x)\]
\[m'(x) = \dfrac{d}{{dx}}\left( {m(x)} \right) = \dfrac{d}{{dx}}\left( {\sin x} \right)\]
Using the derivative formula for \[\sin x\], we have
\[m'(x) = \dfrac{d}{{dx}}\left( {m(x)} \right) = \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\]
Hence, \[m'(x) = \cos x\]
Replacing \[x\] with \[n(x)\], we have
\[m'(n(x)) = \cos (n(x))\]
Using (4), we get
\[m'\left( {n\left( x \right)} \right) = \cos \left( {ax + b} \right) - - - - - - (7)\]
Now, finding \[n'(x)\], we have
\[n'(x) = \dfrac{d}{{dx}}\left( {n\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {ax + b} \right)\]
Now Separating the terms to be differentiated,
\[n'(x) = \dfrac{d}{{dx}}\left( {n\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {ax + b} \right)\]
\[ = \dfrac{d}{{dx}}\left( {ax} \right) + \dfrac{d}{{dx}}\left( b \right)\]
We know, \[\dfrac{d}{{dx}}\left( {ax} \right) = a\] and \[\dfrac{d}{{dx}}\left( b \right) = 0\]
Hence, \[n'(x) = \dfrac{d}{{dx}}\left( {n\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {ax + b} \right)\]
\[ = \dfrac{d}{{dx}}\left( {ax} \right) + \dfrac{d}{{dx}}\left( b \right)\]
\[ = a + 0\]
\[ = a\]
Hence, we get \[n'(x) = a - - - - - - (8)\]
Using (5), (6), (7) and (8), we get
\[\dfrac{d}{{dx}}\left( {mon(x)} \right) = m'(n(x)) \times n'(x)\]
\[\dfrac{d}{{dx}}\left( {\sin \left( {ax + b} \right)} \right) = \cos \left( {ax + b} \right) \times a\]
\[\dfrac{d}{{dx}}\left( {\sin \left( {ax + b} \right)} \right) = a\cos \left( {ax + b} \right) - - - - - - (9)\]
Now, finding \[\dfrac{d}{{dx}}\left( {\cos \left( {ax + b} \right)} \right)\]
Let \[i(x) = \cos x - - - - - - (10)\]
And \[j(x) = cx + d - - - - - - (11)\]
So, \[ioj(x) = i\left( {j\left( x \right)} \right) = \cos \left( {j\left( x \right)} \right) = \cos \left( {cx + d} \right) - - - - - - (12)\]
Using chain rule, \[\dfrac{d}{{dx}}\left( {ioj\left( x \right)} \right) = i'\left( {j\left( x \right)} \right) \times j'(x) - - - - - - (13)\]
First of all, finding \[i'(x)\]
\[i'(x) = \dfrac{d}{{dx}}\left( {i(x)} \right) = \dfrac{d}{{dx}}\left( {\cos x} \right)\]
Using derivative formula for \[\cos x\], we have
\[i'(x) = \dfrac{d}{{dx}}\left( {i(x)} \right) = \dfrac{d}{{dx}}\left( {\cos x} \right)\]
\[ = - \sin x\]
Hence, \[i'(x) = - \sin x\]
Replacing \[x\] with \[j\left( x \right)\], we get
\[i'\left( {j\left( x \right)} \right) = - \sin \left( {j\left( x \right)} \right)\]
Using (11), we get
\[i'\left( {j\left( x \right)} \right) = - \sin \left( {j\left( x \right)} \right) = - \sin \left( {cx + d} \right) - - - - - - - (14)\]
Now, finding \[j'(x)\].
\[j'(x) = \dfrac{d}{{dx}}\left( {j\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {cx + d} \right)\]
Now, separating the terms to be differentiated, we have
\[j'(x) = \dfrac{d}{{dx}}\left( {j\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {cx + d} \right)\]
\[ = \dfrac{d}{{dx}}\left( {cx} \right) + \dfrac{d}{{dx}}\left( d \right)\]
We know, \[\dfrac{d}{{dx}}\left( {cx} \right) = c\] and \[\dfrac{d}{{dx}}\left( d \right) = 0\]
So,
\[j'(x) = \dfrac{d}{{dx}}\left( {j\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {cx + d} \right)\]
\[ = \dfrac{d}{{dx}}\left( {cx} \right) + \dfrac{d}{{dx}}\left( d \right)\]
\[ = c + 0\]
\[ = c\]
Hence, we get \[j'(x) = c - - - - - - (15)\]
Using (12), (13), (14) and (15)
\[\dfrac{d}{{dx}}\left( {ioj\left( x \right)} \right) = i'\left( {j\left( x \right)} \right) \times j'(x)\]
\[\dfrac{d}{{dx}}\left( {\cos \left( {cx + d} \right)} \right) = - \sin \left( {cx + d} \right) \times c\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos \left( {cx + d} \right)} \right) = - c\sin \left( {cx + d} \right) - - - - - - (16)\]
Now, using (9) and (16) in (2), we get
\[\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( {\cos (cx + d) \times \dfrac{d}{{dx}}\left( {\sin (ax + b)} \right)} \right) - \left( {\sin (ax + b) \times \dfrac{d}{{dx}}\left( {\cos (cx + d)} \right)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( {\cos (cx + d) \times a\cos (ax + b)} \right) - \left( {\sin (ax + b) \times \left( { - c\sin (cx + d)} \right)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}\]
Taking out the constant terms outside the brackets,
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( a \right)\left( {\cos (cx + d) \times \cos (ax + b)} \right) - \left( { - c} \right)\left( {\sin (ax + b) \times \sin (cx + d)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( a \right)\left( {\cos (cx + d) \times \cos (ax + b)} \right) + \left( c \right)\left( {\sin (ax + b) \times \sin (cx + d)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}\]
Separating the denominator, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( a \right)\left( {\cos (cx + d) \times \cos (ax + b)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}} + \dfrac{{\left( c \right)\left( {\sin (ax + b) \times \sin (cx + d)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}\]
Cancelling out and Separating the terms, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( a \right)\left( {\cos (ax + b)} \right)}}{{\left( {\cos (cx + d)} \right)}} + \left( c \right)\left( {\sin (ax + b)} \right) \times \dfrac{{\sin (cx + d)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \left( a \right)\left( {\cos (ax + b)} \right) \times \dfrac{1}{{\left( {\cos (cx + d)} \right)}} + \left( c \right)\left( {\sin (ax + b)} \right) \times \dfrac{{\sin (cx + d)}}{{\cos (cx + d)}} \times \dfrac{1}{{\cos (cx + d)}}\]
Now, we know, \[\dfrac{{\sin x}}{{\cos x}} = \tan x\] , \[\dfrac{1}{{\cos x}} = \sec x\]. So, by using this, we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \left( a \right)\left( {\cos (ax + b)} \right) \times \sec \left( {cx + d} \right) + \left( c \right)\left( {\sin (ax + b)} \right) \times \tan \left( {cx + d} \right) \times \sec \left( {cx + d} \right)\]
Hence, derivative of \[\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}\] with respect to \[x\] is \[\left( a \right)\left( {\cos (ax + b)} \right) \times \sec \left( {cx + d} \right) + \left( c \right)\left( {\sin (ax + b)} \right) \times \tan \left( {cx + d} \right) \times \sec \left( {cx + d} \right)\]

Note:
Firstly, we need to see the type of function we are given to differentiate. If the function is composed of two or more functions, always, chain rule is applied. Also, the formula for quotient rule needs to be remembered correctly. We usually put plus signs in between but in quotient rule, it is minus sign. Also, while solving for the composite functions, we need to write the composition of the function carefully.