Question

If f(x) = (cos x)(cos2x)\cdots⋯ (cos nx) then f'(x) + 
$\sf \sum \limits_{r = 1}^{n} \{r \: tan(rx) \}$
 =
1) f(x)
2) 0
3) - f(x)
4) 2f(x) ​

Answer 1

Step-by-step explanation:

Topic :-

Differentiation

Given :-

f(x) = (cosx)(cos2x). . . . (cosnx)

To Find :-

$\sf {f'(x)+\displaystyle \sum_{r=1}^{n}\{rtan(rx)\}f(x)}$

Method :-

We can use Logarithmic function for derivation or we can multiply and divide f(x) by sinx for the simplification and then differentiate it.

We will be using Logarithmic function here.

Solution :-

f(x) = (cosx)(cos2x). . . . (cosnx)

Take 'log' both sides,

log(f(x)) = log((cosx)(cos2x). . . . (cosnx))

We know that,

log(abcd) = log(a) + log(b) + log(c) + log(d)

Applying this,

log(f(x)) = log(cosx) + log(cos2x) +. . . . .+ log(cosnx)

Differentiate both sides,

$\sf {\dfrac{f'(x)}{f(x)}=-\dfrac{sinx}{cosx}-\dfrac{2sin2x}{cos2x}-. . . . . -\dfrac{nsin(nx)}{cosnx}}$

$\sf{f'(x)=\{-tan(x)-2tan(2x)-. . . . -ntan(nx)\}f(x)}$

We can write it as :

$\sf{f'(x)=-\displaystyle \sum_{r=1}^{n}\{rtan(rx)\}f(x)}$

$\sf{f'(x)+\displaystyle \sum_{r=1}^{n}\{rtan(rx)\}f(x)=0}$

Answer :-

So, answer is Zero( 0 ) which is option 2..