Question

Differentiate \[y = {e^{3x}}\sin 4x\] with respect to x.

Answer 1

Hint: The function to be differentiated is a product of two functions of x. Hence, use product rules to differentiate it and then simplify the terms.

Complete step-by-step answer:

We observe that the term to be differentiated is the product of two functions \[{e^{3x}}\] and sin(4x).

We know that to differentiate these terms, we must use the product rule of differentiation.
The product rule of differential calculus states that the differentiation of a product of two functions is the sum of products of one function and the differentiation of the other function and it is given as
follows:
\[(uv)' = uv' + u'v.........(1)\]
where u and v are two functions of x and u’ and v’ are differentiation of u and v with respect to x
respectively.

It is given that,
\[y = {e^{3x}}\sin 4x........(2)\]
We differentiate both sides of the equation (2) to get an expression for \[\dfrac{{dy}}{{dx}}\].
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{e^{3x}}\sin 4x} \right)..........(3)\]
Using the formula in equation (1) in equation (3), we get the following:
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = {e^{3x}}\dfrac{d}{{dx}}\left( {\sin 4x} \right) + \sin 4x\dfrac{d}{{dx}}\left({{e^{3x}}} \right)..........(3)\]


We know that differentiation of sin(ax) is a.cos(ax) and the differentiation of \[{e^{ax}}\] is
\[a{e^{ax}}\]. Using these formulas to simplify equation (3), we get:
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = {e^{3x}}.4\cos 4x + \sin 4x.3{e^{3x}}\]
Taking \[{e^{3x}}\] as a common term, we get the final expression as:
$\Rightarrow$ \[\dfrac{{dy}}{{dx}} = {e^{3x}}(4\cos 4x + 3\sin 4x)\]
Hence, the answer is \[{e^{3x}}(4\cos 4x + 3\sin 4x)\].

Note: You can easily forget the constant term when differentiating sin(4x) and \[{e^{3x}}\] and the you might get the final answer as \[{e^{3x}}(\cos 4x + \sin 4x)\], which is wrong. This question is an example for application of the product rule of differentiation.