Question

Find the derivative of the function$\dfrac{d}{{dx}}\left[ {{e^{ax}}Cos\left( {bx + c} \right)} \right]$?
A) ${e^{ax}}\left( {a\operatorname{Cos} \left( {bx + c} \right) + b\operatorname{Sin} \left( {bx + c} \right)} \right)$
B) ${e^{ax}}\left( {a\operatorname{Cos} \left( {bx + c} \right) - b\operatorname{Sin} \left( {bx + c} \right)} \right)$
C) ${e^{ax}}\left( {\operatorname{Cos} \left( {bx + c} \right) - b\operatorname{Sin} \left( {bx + c} \right)} \right)$
D) None of the above

Answer 1

Hint: Here we have to differentiate $\dfrac{d}{{dx}}\left[ {{e^{ax}}Cos\left( {bx + c} \right)} \right]$ with respect to $x$. We solve this by using the product function rule of differentiation. Let $f$ and$g$ be two functions. Derivative of product of two functions is given by the following product function rule:
$\dfrac{d}{{dx}}\left[ {f\left( x \right).g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right).g\left( x \right) + f\left( x \right).\dfrac{d}{{dx}}g\left( x \right)$
I.e. Derivative of product of two functions = [(Derivative of the first function) × (Second function) + (Derivative of the Second function) × (First function)].
We can also write product function rule as: ${\left( {uv} \right)'} = {u'}v + u{v'}$ (Here $'$ means derivative, $f\left( x \right) = u$ and $g\left( x \right) = v$). ${\left( {uv} \right)'} = {u'}v + u{v'}$ is referred to as the Leibnitz rule for differentiating the product of functions or the product rule.
Formula: $\dfrac{d}{{dx}}\left[ {f\left( x \right).g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right).g\left( x \right) + f\left( x \right).\dfrac{d}{{dx}}g\left( x \right)$

Complete step-by-step answer:
Given $\dfrac{d}{{dx}}\left[ {{e^{ax}}Cos\left( {bx + c} \right)} \right]$
Let $y = {e^{ax}}.\operatorname{Cos} \left( {bx + c} \right)$
Differentiate the above function with respect to $x$ using the product function rule:
Product function rule: ${\left( {uv} \right)'} = {u'}v + u{v'}$
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{e^{ax}}.\operatorname{Cos} \left( {bx + c} \right) + {e^{ax}}\dfrac{d}{{dx}}\operatorname{Cos} \left( {bx + c} \right)$
Steps to solve R.H.S.
\[ = \operatorname{Cos} \left( {bx + c} \right)\dfrac{d}{{dx}}{e^{ax}} + {e^{ax}}\dfrac{d}{{dx}}\operatorname{Cos} \left( {bx + c} \right)\]
Here to find derivative will use $\dfrac{d}{{dx}}{e^x} = {e^x}$ and $\dfrac{d}{{dx}}\left( {\operatorname{Cos} x} \right) = - \operatorname{Sin} x$.
\[ = \operatorname{Cos} \left( {bx + c} \right).a{e^{ax}} + {e^{ax}}.\left( { - \operatorname{Sin} \left( {bx + c} \right).b} \right)\]
\[ = a{e^{ax}}.\operatorname{Cos} \left( {bx + c} \right) - b{e^{ax}}\operatorname{Sin} \left( {bx + c} \right)\]
Take ${e^{ax}}$ common
\[ = {e^{ax}}\left( {a\operatorname{Cos} \left( {bx + c} \right) - b\operatorname{Sin} \left( {bx + c} \right)} \right)\]
Hence, the correct option is 2.
So, the correct answer is “Option 2”.

Note: In simple words we can describe the product function rule as the derivative of the product of two functions is equal to the second function times the derivative of the first function plus the first function times the derivative of the second function. The most important point to remember about the product function rule is that the derivative of the product of two functions is not the simple product of their separate derivatives i.e. ${\left( {uv} \right)'} \ne {u'} + {v'}$. There are three rules that are generally applicable, and depend on the structure of the function we are differentiating. These are the product, quotient, and chain rules.