Question

Differentiate with respect to x
$\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}$

Answer 1

We have to solve this differentiation using quotient rule. We separately solve for derivatives of numerator and denominator using chain rule and then substitute in the formula for quotient rule. Then separating the terms we try to form trigonometric forms in the final answer.

• Quotient rule states that differentiation of $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}$ where $u'$is differentiation of u with respect to x and $v'$is differentiation of v with respect to x.
• Chain rule which will be used within the differentiation is given as $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$ where $f'$denotes differentiation of function f with respect to x and $g'$denotes differentiation of function g with respect to x.

Complete step-by-step answer:
We write the differentiation as $\dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right)$.
Now comparing with $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right)$, we get $u = \sin (ax + b),v = \cos (cx + d)$
First we find the values of $u'$and $v'$using chain rule.
Since, $u = \sin (ax + b)$
$\Rightarrow u' = \dfrac{{d(u)}}{{dx}} = \dfrac{d}{{dx}}(\sin (ax + b))$
According to chain rule $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$
Substitute the value of $f(x) = \sin (g(x)),g(x) = (ax + b)$
Then $f'(x) = \dfrac{{df(x)}}{{dx}} = \dfrac{{d\sin (g(x))}}{{dx}} = \cos (g(x))$ and $g'(x) = \dfrac{{dg(x)}}{{dx}} = \dfrac{{d(ax + b)}}{{dx}} = a$
$\dfrac{d}{{dx}}\left[ {\sin (ax + b)} \right] = \cos (ax + b).a$
$\Rightarrow u' = a\cos (ax + b)$ … (1)
Since, $v = \cos (cx + d)$
$\Rightarrow v' = \dfrac{{d(v)}}{{dx}} = \dfrac{d}{{dx}}(\cos (cx + d))$
According to chain rule $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$
Substitute the value of $f(x) = \cos (g(x)),g(x) = (cx + d)$
Then $f'(x) = \dfrac{{df(x)}}{{dx}} = \dfrac{{d\cos (g(x))}}{{dx}} = - \sin (g(x))$ and $g'(x) = \dfrac{{dg(x)}}{{dx}} = \dfrac{{d(cx + d)}}{{dx}} = c$
$\dfrac{d}{{dx}}\left[ {\cos (cx + d)} \right] = - \sin (cx + d).c$
$\Rightarrow v' = - c\sin (cx + d)$ … (2)
We solve$\dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right)$ using quotient rule which is $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{vu' - uv'}}{{{v^2}}}$
Here, $u = \sin (ax + b),v = \cos (cx + d)$and $u' = a\cos (ax + b),v' = - c\sin (cx + d)$
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{\left( {\cos (cx + d)} \right)\left( {a\cos (ax + b)} \right) - \left( {\sin (ax + b)} \right)\left( { - c\sin (cx + d)} \right)}}{{{{\left( {\cos (cx + d)} \right)}^2}}}$
Multiplying the brackets in the numerator
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{a\left( {\cos (ax + b)\cos (cx + d)} \right) + c\left( {\sin (ax + b)\sin (cx + d)} \right)}}{{{{\cos }^2}(cx + d)}}$
Now we separate the terms in the numerator
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{a\left( {\cos (ax + b)\cos (cx + d)} \right)}}{{\cos (cx + d)\cos (cx + d)}} + \dfrac{{c\left( {\sin (ax + b)\sin (cx + d)} \right)}}{{\cos (cx + d)\cos (cx + d)}}$
Cancelling out the same terms from numerator and denominator.
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = \dfrac{{a\left( {\cos (ax + b)} \right)}}{{\cos (cx + d)}} + \dfrac{{c\left( {\sin (ax + b)\sin (cx + d)} \right)}}{{\cos (cx + d)\cos (cx + d)}}$
Now we separate the terms
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = a\cos (ax + b).\dfrac{1}{{\cos (cx + d)}} + c\sin (ax + b).\dfrac{{\sin (cx + d)}}{{\cos (cx + d)}}.\dfrac{1}{{\cos (cx + d)}}$
Substituting the values of $\dfrac{1}{{\cos (cx + d)}} = \sec (cx + d),\dfrac{{\sin (cx + d)}}{{\cos (cx + d)}} = \tan (cx + d)$
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}} \right) = a\cos (ax + b)\sec (cx + d) + c\sin (ax + b)\tan (cx + d)\sec (cx + d)$
Thus differentiation of $\dfrac{{\sin (ax + b)}}{{\cos (cx + d)}}$ with respect to x is $a\cos (ax + b)\sec (cx + d) + c\sin (ax + b)\tan (cx + d)\sec (cx + d)$

Note: Students are likely to make mistake of multiplying both numerator and denominator by 2 to make use of the formula which will convert the numerator i.e.$2\sin A\sin B = \cos (A - B) - \cos (A + B)$ and $2\cos A\cos B = \cos (A + B) + \cos (A - B)$
Students are advised not to operate in such a way because this makes our angles inside the bracket even more complicated.