Question

Evaluate $\dfrac{d}{{dx}}\left( {\dfrac{{{e^{ax}}}}{{\sin \left( {b + c} \right)}}} \right)$.
A. $\dfrac{{{a^{ax}}\left[ {a\sin \left( {bx + c} \right) + b\cos \left( {bx + c} \right)} \right]}}{{{{\sin }^2}\left( {bx + c} \right)}}$
B. $\dfrac{{{a^{ax}}\left[ {a\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right]}}{{\sin \left( {bx + c} \right)}}$
C. $\dfrac{{{e^{ax}}\left[ {a\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right]}}{{{{\sin }^2}\left( {bx + c} \right)}}$
D. None of these

Answer 1

Hint: Given question is related to derivative. To find the derivative of $\dfrac{{{e^{ax}}}}{{\sin \left( {bx + c} \right)}}$, we will apply the quotient formula.

Formula Used:
Quotient formula: $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$
$\dfrac{d}{{dx}}{e^{mx}} = m{e^{mx}}$ Where $m$ is a constant.
Chain rule: $\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)$
$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$\dfrac{d}{{dx}}\left( {mx} \right) = m$ Where $m$ is a constant.

Complete step by step solution:
Given that, $\dfrac{d}{{dx}}\left( {\dfrac{{{e^{ax}}}}{{\sin \left( {b + c} \right)}}} \right)$
Apply the quotient formula $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$
$\begin{array}{l} = \dfrac{d}{{dx}}\left( {\dfrac{{{e^{ax}}}}{{\sin \left( {bx + c} \right)}}} \right)\\ = \dfrac{{\sin \left( {bx + c} \right)\dfrac{d}{{dx}}{e^{ax}} - {e^{ax}}\dfrac{d}{{dx}}\sin \left( {bx + c} \right)}}{{{{\sin }^2}\left( {bx + c} \right)}}\end{array}$
$ = \dfrac{{\sin \left( {bx + c} \right)\dfrac{d}{{dx}}{e^{ax}} - {e^{ax}}\dfrac{d}{{dx}}\sin \left( {bx + c} \right)}}{{{{\sin }^2}\left( {bx + c} \right)}}$
Now applying $\dfrac{d}{{dx}}{e^{mx}} = m{e^{mx}}$and chain rule $\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)$
$ = \dfrac{{a{e^{ax}}\sin \left( {bx + c} \right) - {e^{ax}}\cos \left( {bx + c} \right)\dfrac{d}{{dx}}\left( {bx + c} \right)}}{{{{\sin }^2}\left( {bx + c} \right)}}$
Now applying $\dfrac{d}{{dx}}\left( {f\left( x \right) \pm g\left( x \right)} \right) = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$
$ = \dfrac{{a{e^{ax}}\sin \left( {bx + c} \right) - {e^{ax}}\cos \left( {bx + c} \right)\left[ {\dfrac{d}{{dx}}\left( {bx} \right) + \dfrac{d}{{dx}}c} \right]}}{{{{\sin }^2}\left( {bx + c} \right)}}$
Now applying $\dfrac{d}{{dx}}\left( {mx} \right) = m$ and $\dfrac{d}{{dx}}m = 0$
$ = \dfrac{{a{e^{ax}}\sin \left( {bx + c} \right) - {e^{ax}}\cos \left( {bx + c} \right) \cdot b}}{{{{\sin }^2}\left( {bx + c} \right)}}$
$ = \dfrac{{a{e^{ax}}\sin \left( {bx + c} \right) - b{e^{ax}}\cos \left( {bx + c} \right)}}{{{{\sin }^2}\left( {bx + c} \right)}}$
$ = \dfrac{{{e^{ax}}\left[ {a\sin \left( {bx + c} \right) - b\cos \left( {bx + c} \right)} \right]}}{{{{\sin }^2}\left( {bx + c} \right)}}$

Option ‘C’ is correct

Note: This question can be solved by using quotient formula and chain rule. First apply the quotient formula on the given expression as it is in fractional form. To find $\dfrac{d}{{dx}}\sin \left( {bx + c} \right)$,we will apply the chain rule as b is multiplied with x.