   Question Answers

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

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.
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