Question

If $y={{e}^{ax}}\cdot \cos \left( bx+c \right)$, then find $\dfrac{dy}{dx}$

Answer 1

Hint: In this question we have been given an expression for which we have to find its derivative. Since there is no direct way to differentiate the given term, we will use the product rule which is given as $\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. we will then use the chain rule during differentiating the terms. and get the required solution.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow y={{e}^{ax}}\cdot \cos \left( bx+c \right)$
we have to find the derivative of this equation therefore; it can be written as:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}{{e}^{ax}}\cdot \cos \left( bx+c \right)$
On using the product rule $\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ by considering $u={{e}^{ax}}$ and $v=\cos \left( bx+c \right)$, we get:
$\Rightarrow \dfrac{dy}{dx}={{e}^{ax}}\cdot \dfrac{d}{dx}\cos \left( bx+c \right)+\cos \left( bx+c \right)\dfrac{d}{dx}{{e}^{ax}}$
Now we know that $\dfrac{d}{dx}\cos x=-\sin x$ and $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$ therefore, on substituting and using the chain rule, we get:
\[\Rightarrow \dfrac{dy}{dx}=-{{e}^{ax}}\sin \left( bx+c \right)\dfrac{d}{dx}\left( bx+c \right)+\cos \left( bx+c \right){{e}^{ax}}\dfrac{d}{dx}ax\]
Now we know that $\dfrac{d}{dx}kx=k$ therefore, we get:
\[\Rightarrow \dfrac{dy}{dx}=-{{e}^{ax}}\sin \left( bx+c \right)\times b+\cos \left( bx+c \right){{e}^{ax}}\times a\]
On multiplying and rearranging the terms, we get:
\[\Rightarrow \dfrac{dy}{dx}=a{{e}^{ax}}\cos \left( bx+c \right)-b{{e}^{ax}}\sin \left( bx+c \right)\], which is the required derivative.

Note: To do these types of questions all the basic derivative formulas should be remembered.
In this question we have used the product formula, there also exists the division formula which is:
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}$ which should be remembered to do the division sums.
It is to be remembered that integration and derivative are the inverse of each other, if the derivative of $a$ is $b$, then the integration of $b$ will be $a$. If there is a term which has more than to functions of derivatives which means that it is the form of $f(g(x))$, then the chain rule has to be used which is $F'(x)=f'(g(x))g'(x)$, this should be done till the point the equation is in the simple format. It is to be remembered that the property of chain rule is only unique to differentiation and not integration and therefore the chain rule should not be used in integration questions. Also, the product rule is different for integration questions.