The minimum value of $$2^{\sin x} + 2^{\cos x}$$ is:

We want the least possible value of the function

$$f(x)=2^{\sin x}+2^{\cos x},\qquad x\in\mathbb R.$$

Because the expression is periodic with period $$2\pi,$$ it is enough to study one full period, for example $$x\in[0,2\pi).$$ To locate extrema we differentiate. The basic differentiation formula we need is

$$\dfrac{d}{dx}\bigl(2^{u(x)}\bigr)=2^{u(x)}\ln 2\;u'(x).$$

Applying it first to $$2^{\sin x}$$ and then to $$2^{\cos x}$$ gives

$$\frac{d}{dx}\bigl(2^{\sin x}\bigr)=2^{\sin x}\ln 2\;\cos x,$$

$$\frac{d}{dx}\bigl(2^{\cos x}\bigr)=2^{\cos x}\ln 2\;(-\sin x).$$

So the derivative of the whole function is

$$f'(x)=2^{\sin x}\ln 2\;\cos x-2^{\cos x}\ln 2\;\sin x.$$

Factorising the common factor $$\ln 2$$ (which is always positive) we have

$$f'(x)=\ln 2\Bigl(\cos x\;2^{\sin x}-\sin x\;2^{\cos x}\Bigr).$$

The critical points occur when the bracket equals zero, i.e.

$$\cos x\;2^{\sin x}-\sin x\;2^{\cos x}=0.$$

Re-arranging gives the necessary condition

$$\cos x\;2^{\sin x}=\sin x\;2^{\cos x}.$$

Assuming $$\sin x\neq 0$$ and $$\cos x\neq 0$$ (we shall look at those special cases later) we divide by the positive quantities and obtain

$$\frac{\cos x}{\sin x}=2^{\cos x-\sin x}.$$

The left side is $$\cot x,$$ the right side is $$2^{\cos x-\sin x}.$$ This transcendental equation is satisfied when the two exponents of 2 are equal, that is when

$$\sin x=\cos x.$$

Using the identity $$\sin x=\cos x\; \Longrightarrow\; x=\frac{\pi}{4}+k\pi,\;k\in\mathbb Z,$$ we get two types of points in one period:

1. $$x=\frac{\pi}{4}$$ (both sine and cosine positive),
2. $$x=\frac{5\pi}{4}$$ (both sine and cosine negative).

Next we check the endpoints where either sine or cosine is zero, because there the derivative formula involved division by those terms:

a) $$x=0:\;\sin x=0,\;\cos x=1\;\Longrightarrow\;f(0)=2^{0}+2^{1}=1+2=3.$$

b) $$x=\frac{\pi}{2}:\;\sin x=1,\;\cos x=0\;\Longrightarrow\;f\!\left(\frac{\pi}{2}\right)=2^{1}+2^{0}=2+1=3.$$

c) $$x=\pi:\;\sin x=0,\;\cos x=-1\;\Longrightarrow\;f(\pi)=2^{0}+2^{-1}=1+\frac12=\frac32=1.5.$$

d) $$x=\frac{3\pi}{2}:\;\sin x=-1,\;\cos x=0\;\Longrightarrow\;f\!\left(\frac{3\pi}{2}\right)=2^{-1}+2^{0}=\frac12+1=1.5.$$

Now we evaluate the function at the two critical points where $$\sin x=\cos x.$$

For $$x=\frac{\pi}{4}$$ we have $$\sin x=\cos x=\frac1{\sqrt2},$$ so

$$f\!\left(\frac{\pi}{4}\right)=2^{\frac1{\sqrt2}}+2^{\frac1{\sqrt2}}=2\cdot2^{\frac1{\sqrt2}}=2^{1+\frac1{\sqrt2}}\approx3.266.$$

For $$x=\frac{5\pi}{4}$$ we have $$\sin x=\cos x=-\frac1{\sqrt2},$$ so

$$f\!\left(\frac{5\pi}{4}\right)=2^{-\frac1{\sqrt2}}+2^{-\frac1{\sqrt2}}=2\cdot2^{-\frac1{\sqrt2}}=2^{1-\frac1{\sqrt2}}\approx1.225.$$

Comparing all the values we have collected

$$\begin{aligned} f\!\left(\frac{5\pi}{4}\right)&=2^{1-\frac1{\sqrt2}}\approx1.225,\\[4pt] f(\pi)=f\!\left(\frac{3\pi}{2}\right)&=1.5,\\[4pt] f(0)=f\!\left(\frac{\pi}{2}\right)&=3,\\[4pt] f\!\left(\frac{\pi}{4}\right)&\approx3.266, \end{aligned}$$

and note that all other points in the interval $$[0,2\pi)$$ lie between these tested points. Hence the smallest of all these numbers, and therefore the global minimum of $$f(x),$$ is

$$f_{\min}=2^{1-\frac1{\sqrt2}}.$$

Thus the minimum value of $$2^{\sin x}+2^{\cos x}$$ is $$2^{1-\frac1{\sqrt2}}.$$

Hence, the correct answer is Option D.