Correct Answer - Option 1 : -1
Concept:
- log m + log n = log mn
- sin x/cos x = tan x
- \(\rm \frac{d}{dx} log\ x = \frac{1}{x}\)
- \(\rm \frac{d}{dx} cos x = -sin x\)
Calculation:
y = cos x ⋅ cos 4x ⋅ cos 8x
Taking log on both side
⇒ log y = log (cos x ⋅ cos 4x ⋅ cos 8x)
⇒ log y = log cos x + log cos 4x + log cos 8x
Differentiating both side with respect to x
⇒ \(\frac{1}{y}\frac{dy}{dx} = \frac{-sin\ x}{cos\ x}+\frac{-4sin\ 4x}{cos\ 4 x}+\frac{-8sin\ 8x}{cos\ 8x}\)
⇒ \(\frac{1}{y}\frac{dy}{dx}\) = -1[tan x + tan 4x + tan 8x]
Now putting \(\rm x = \frac{π}{4}\)
⇒ \(\frac{1}{y}\frac{dy}{dx}\) = -1[\(tan\ \frac{π}{4}+tan \frac{4π}{4}+tan\ \frac{8π}{4}\)]
⇒ \(\frac{1}{y}\frac{dy}{dx}\) = -[1 + tan π + tan 2π] [∵ tan(π/4) = 1]
⇒ \(\frac{1}{y}\frac{dy}{dx}\) = -1 (∵ tan nπ = 0)
∴ \(\rm \frac{1}{y}\frac{dy}{dx}\) at \(x = \frac{π}{4}\) equal to is -1.
