Question

Find $\dfrac{{dy}}{{dx}}$, when $y = \cos x\cos 2x\cos 3x$.

Answer 1

To solve this problem, we will take logarithmic functions on both sides of $y$. Then we will differentiate both sides with respect to $x$. To differentiate both the sides we will use the basic theories of differentiation like chain rule. Further keeping $\dfrac{{dy}}{{dx}}$ on the left hand side of the equation and taking all the other terms on the right hand side and simplifying, we will get the required answer.

Complete step-by-step solution:
Given equation is $y = \cos x\cos 2x\cos 3x$.
Now, taking logarithmic function on both the sides of the equation, we get,
$\Rightarrow \log y = \log \left( {\cos x\cos 2x\cos 3x} \right)$
We know, $\log \left( {x.y.z} \right) = \log x + \log y + \log z$.
Using this property of logarithmic function in the given equation, we get,
$\Rightarrow \log y = \log \left( {\cos x} \right) + \log \left( {\cos 2x} \right) + \log \left( {\cos 3x} \right)$
Now, differentiating both the sides of the equation with respect to $x$, we get,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}}.\dfrac{{d\left( {\cos x} \right)}}{{dx}} + \dfrac{1}{{\cos 2x}}.\dfrac{{d\left( {\cos 2x} \right)}}{{dx}} + \dfrac{1}{{\cos 3x}}.\dfrac{{d\left( {\cos 3x} \right)}}{{dx}}$
[Using, $\dfrac{{d\left( {\log x} \right)}}{{dx}} = \dfrac{1}{x}$ and chain rule]
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{{\cos x}}.\left( { - \sin x} \right) + \dfrac{1}{{\cos 2x}}.\left( { - \sin 2x} \right).\dfrac{{d\left( {2x} \right)}}{{dx}} + \dfrac{1}{{\cos 3x}}.\left( { - \sin 3x} \right).\dfrac{{d\left( {3x} \right)}}{{dx}}$
[Using, $\dfrac{{d\left( {\cos x} \right)}}{{dx}} = - \sin x$ and chain rule]
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = - \dfrac{{\sin x}}{{\cos x}} - \dfrac{{\sin 2x}}{{\cos 2x}}.2 - \dfrac{{\sin 3x}}{{\cos 3x}}.3$
Now, simplifying the terms, we get,
$\Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = - \tan x - 2.\tan 2x - 3\tan 3x$
[Using, $\dfrac{{\sin x}}{{\cos x}} = \tan x$]
Multiplying $y$ on both the sides, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = y\left[ { - \tan x - 2.\tan 2x - 3\tan 3x} \right]$
Taking negative common on right hand side of the equation, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - y\left[ {\tan x + 2.\tan 2x + 3\tan 3x} \right]$
Given, $y = \cos x\cos 2x\cos 3x$.
So, substituting this in the equation, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \cos x\cos 2x\cos 3x\left[ {\tan x + 2.\tan 2x + 3\tan 3x} \right]$.

Note: We can also solve this problem by using chain rule and the formula of differentiation of products. On right hand of the equation, taking two of the cosine terms together and the other term separately, like, taking $\cos x$ as one term and $\left( {\cos 2x\cos 3x} \right)$ as another term and using the formula of differentiation of products which is $\dfrac{{d\left( {uv} \right)}}{{dx}} = \dfrac{{du}}{{dx}}.v + u.\dfrac{{dv}}{{dx}}$. Then using chain rule and differentiating the two cosine terms $\left( {\cos 2x\cos 3x} \right)$ by the formula of differentiation of products, we will get the required value.