Question

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

Answer 1

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.

Answer 2

Answer :-

option (2) 0 is correct

Additional Information :-

$\boxed{\boxed{\begin{minipage}{4.2cm}\circ\sf\:\ln 1=0\\\circ\sf\:\ln e=1\\\circ\sf\:\ln x=y\leftrightarrow e^y=x\\\circ\sf\:e^{\ln x}=x,\:x>0\\\circ\sf\:\ln(e^x)=x,\:x\in\mathbb{\large R}\\\circ\sf\:\ln(xy)=\ln x+\ln y\\\circ\sf\:\ln(x/y)=\ln x-\ln y\\\circ\sf\:\ln(x^r)=r\ln x\\\circ\sf\:\ln x=log_e\:x\end{minipage}}}$

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\ \circ \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \circ \; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\ \circ\; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \circ \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \circ \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \circ\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \circ\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\ \circ\;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\ \circ\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \circ\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \circ\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx\end{minipage}}}$