Question

What is the derivative of \[\sin 3x\]?

Answer 1

Hint: Here, we are asked to find the derivative of \[\sin 3x\] with respect to \[x\]. We will first let the equation be equal to \[y\]. As we can see it consists of two functions. One of the functions is the trigonometric function whereas the other is the algebraic function. So, we will use the chain rule of differentiation to find the derivative of the given function.

Complete step-by-step answer:
Let,
\[y = \sin 3x\]
Now we will differentiate both the sides of the above equation with respect to \[x\]. So, we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sin 3x} \right)}}{{dx}}\]
Since, \[\sin 3x\] contains two functions that are trigonometric and algebraic, so we will use the chain rule of differentiation.
So, according to the chain rule, as the angle in \[\sin 3x\] is \[3x\] so we will first differentiate it with respect to \[3x\] and then we will differentiate \[3x\] with respect to \[x\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\left( {\sin 3x} \right)}}{{d\left( {3x} \right)}} \times \dfrac{{d\left( {3x} \right)}}{{dx}}\]
Now, we know differentiation of \[\sin x\] with respect to \[x\] is \[\cos x\] i.e., \[\dfrac{{d\sin x}}{{dx}} = \cos x\].
Also, differentiation of \[3x\] with respect to \[x\] is \[3\]. So, we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {3x} \right) \times 3\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3\cos 3x\]
So, the correct answer is “\[3\cos 3x\]”.

Note: For the solution we have used the chain rule of differentiation but we can also solve this using one simple shortcut or rather a formula that is;
\[\dfrac{{d\sin \left( {ax + b} \right)}}{{dx}} = a\cos \left( {ax + b} \right)\].
So, if we compare the given function that is \[\sin 3x\] with \[\sin \left( {ax + b} \right)\], then we get \[a = 3,b = 0\].
So, using the above formula we get;
\[ \Rightarrow \dfrac{{d\sin 3x}}{{dx}} = 3\cos 3x\]
Another point to note is that if in the question it is asked that what is the derivative of \[\sin 3x\] with respect to \[y\], then it will be zero because in that case \[\sin 3x\] will behave as constant and we know that the differentiation of constant is equal to zero.