Answer

Verified

450.9k+ views

Hint: In this problem we have to convert the product of three trigonometric function in the form of addition by applying log properties i.e log(axbxc) = log a +log b + log c .

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]$ .

Note: We can also use the formula

$\dfrac{{d({y_1}.{y_2})}}{{dx}}= {y_1}\dfrac{{d({y_2})}}{{dx}} + {y_2}\dfrac{{d({y_1})}}{{dx}}$. 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.

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]$ .

Note: We can also use the formula

$\dfrac{{d({y_1}.{y_2})}}{{dx}}= {y_1}\dfrac{{d({y_2})}}{{dx}} + {y_2}\dfrac{{d({y_1})}}{{dx}}$. 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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

Difference Between Plant Cell and Animal Cell

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

One cusec is equal to how many liters class 8 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How many crores make 10 million class 7 maths CBSE

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