Differentiate the function 2^cos^2 x with respect to x.

The correct answer is Option (2) → $-2^{\cos^2 x} \cdot \log 2 (\sin 2x)$ ##

Let $y = 2^{\cos^2 x}$

Taking log on both sides, we get

$\log y = \log 2^{\cos^2 x} = \cos^2 x \cdot \log 2 \quad [∵\log m^n = n \log m]$

On differentiating w.r.t. $x$, we get

$\frac{d}{dy} \log y \cdot \frac{dy}{dx} = \frac{d}{dx} (\log 2 \cdot \cos^2 x)$

$\Rightarrow \frac{1}{y} \frac{dy}{dx} = \log 2 \frac{d}{dx} (\cos x)^2 \Rightarrow \frac{1}{y} \frac{dy}{dx} = \log 2 \cdot [2 \cos x] \cdot \frac{d}{dx} \cos x$

$= \log 2 \cdot 2 \cos x \cdot (-\sin x) \quad \left[ ∵\frac{d}{dx} \cos x = -\sin x \right]$

$= \log 2 \cdot [-(\sin 2x)] \quad [∵\sin 2x = 2 \sin x \cdot \cos x]$

$\Rightarrow \frac{dy}{dx} = -y \cdot \log 2 (\sin 2x)$

$∴\frac{dy}{dx} = -2^{\cos^2 x} \cdot \log 2 (\sin 2x) \quad [∵y = 2^{\cos^2 x}]$