Question

# Differentiate the function with respect to x.$\cos x.\cos 2x.\cos 3x$

Let, $y = \cos x.\cos 2x.\cos 3x$ . Since their terms are in multiplication which we don’t want and we can get rid of this by taking logarithmic both sides because we know that $\log (m.n) = \log (m) + \log (n)$ .
$\ y = \cos x.\cos 2x.\cos 3x \\ \Rightarrow \log y = \log (\cos x) + \log (\cos 2x) + \log (\cos 3x) \\ \$
As the problem statement said, we’ll now differentiate both sides using chain rule and formula$\dfrac{{d(\log x)}}{{dx}} = \dfrac{1}{x}$
$\ \log y = \log (\cos x) + \log (\cos 2x) + \log (\cos 3x) \\ \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) \\ \Rightarrow \dfrac{{dy}}{{dx}} = y[ - \dfrac{{\sin x}}{{\cos x}} - 2\dfrac{{\sin 2x}}{{\cos 2x}} - 3\dfrac{{\sin 3x}}{{\cos 3x}}] \\ \Rightarrow \dfrac{{dy}}{{dx}} = y[ - \tan x - 2\tan 2x - 3\tan 3x] \\ \Rightarrow \dfrac{{dy}}{{dx}} = - \cos x.\cos 2x.\cos 3x.[\tan x + 2\tan 2x + 3\tan 3x] \\ \$
Hence, the required differentiation of the given function is $- \cos x.\cos 2x.\cos 3x.[\tan x + 2\tan 2x + 3\tan 3x]$ .
$\dfrac{{d({y_1}.{y_2})}}{{dx}}= {y_1}\dfrac{{d({y_2})}}{{dx}} + {y_2}\dfrac{{d({y_1})}}{{dx}}$. But it’ll make the solution lengthy because we have three functions in the multiplication and we need to use the formula two times. In differential calculus, we often use this hack of taking logs to get rid of long solutions.