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

Answer

Verified

380.4k+ views

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.

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.

Recently Updated Pages

Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts

The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE