Question

Differentiate w.r.t x?
\[\cos x \cdot \cos 2x \cdot \cos 3x\]

Answer 1

Hint: Here we have to differentiate the given equation with respect to \[x\]. For that, we will take the logarithm of the given trigonometric function. We will split the terms using the properties of logarithmic function. Then we will differentiate each term properly and we will simplify the like terms in the equation. From there, we will get the result of differentiation of the given expression.

Formula used:
We will use the multiplication property of the logarithmic function, \[\log \left( {a \cdot b \cdot c} \right) = \log a + \log b + \log c\].

Complete step-by-step answer:
Let the given expression \[\cos x \cdot \cos 2x \cdot \cos 3x\] be \[y\] i.e.
\[y = \cos x \cdot \cos 2x \cdot \cos 3x\]
We will take log on sides of equation.
Thus, the equation becomes.
\[ \Rightarrow \log y = \log \left( {\cos x \cdot \cos 2x \cdot \cos 3x} \right)\]
We will use the logarithmic property, \[\log \left( {a \cdot b \cdot c} \right) = \log a + \log b + \log c\], to simplify the above expression. Thus,
\[ \Rightarrow \log y = \log \cos x + \log \cos 2x + \log \cos 3x\]
We will differentiate both sides of the equation with respect to \[x\].
\[ \Rightarrow \dfrac{{d\log y}}{{dx}} = \dfrac{{d\left( {\log \cos x + \log \cos 2x + \log \cos 3x} \right)}}{{dx}}\]
\[ \Rightarrow \dfrac{{d\log y}}{{dx}} = \dfrac{{d\log \cos x}}{{dx}} + \dfrac{{d\log \cos 2x}}{{dx}} + \dfrac{{d\log \cos 3x}}{{dx}}\]
Now, we will differentiate each term one by one.
We know the derivative of \[\log x\] is equal to \[\dfrac{1}{x}\].
So, differentiating the equation, we get
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}} \times - \sin x + \dfrac{1}{{\cos 2x}} \times - 2\sin 2x + \dfrac{1}{{\cos 3x}} \times - 3\sin 3x\]
Simplifying the terms further, we get
\[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = - \tan x - 2\tan 2x - 3\tan 3x\]
Multiplying \[y\] on both sides, we get
\[ \Rightarrow y \times \dfrac{1}{y}\dfrac{{dy}}{{dx}} = y\left( { - \tan x - 2\tan 2x - 3\tan 3x} \right)\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\left( { - \tan x - 2\tan 2x - 3\tan 3x} \right)\]
Substituting \[y = \cos x \cdot \cos 2x \cdot \cos 3x\] in the equation, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \cos x \cdot \cos 2x \cdot \cos 3x\left( { - \tan x - 2\tan 2x - 3\tan 3x} \right)\]
Simplifying the equation further, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \cos x \cdot \cos 2x \cdot \cos 3x\left( { + \tan x + 2\tan 2x + 3\tan 3x} \right)\]
This is the required differentiation of \[\cos x \cdot \cos 2x \cdot \cos 3x\] with respect to \[x\].

Note: For solving this question, we need to know the concept of differentiation and basic functions like trigonometric function, exponential function, logarithmic function etc. Differentiation is defined as the process to find a function that gives output of rate of change of one variable with respect to another.