Let $$f(x) = \sin^4 x + \cos^4 x$$. Then, $$f$$ is an increasing function in the interval:

We have the function

$$f(x)=\sin^{4}x+\cos^{4}x.$$

To find where $$f(x)$$ is increasing, we must examine the sign of its first derivative. The rule we will use is the power rule combined with the chain rule: if $$g(x)=\big(h(x)\big)^n,$$ then $$g'(x)=n\,\big(h(x)\big)^{\,n-1}\,h'(x).$$

First differentiate the two terms separately.

For the sine term:

$$\frac{d}{dx}\bigl(\sin^{4}x\bigr)=4\sin^{3}x\cdot\frac{d}{dx}(\sin x)=4\sin^{3}x\cos x.$$

For the cosine term, note that $$\frac{d}{dx}(\cos x)=-\sin x$$, so

$$\frac{d}{dx}\bigl(\cos^{4}x\bigr)=4\cos^{3}x\cdot\frac{d}{dx}(\cos x)=4\cos^{3}x(-\sin x)=-4\cos^{3}x\sin x.$$

Adding the two derivatives gives

$$f'(x)=4\sin^{3}x\cos x-4\cos^{3}x\sin x.$$

Both terms contain the common factor $$4\sin x\cos x,$$ so we factor it out:

$$f'(x)=4\sin x\cos x\bigl(\sin^{2}x-\cos^{2}x\bigr).$$

The derivative therefore breaks naturally into three factors:

$$\sin x,\qquad\cos x,\qquad\text{and}\qquad\sin^{2}x-\cos^{2}x.$$

To decide the sign of $$f'(x)$$ we examine each factor on the interval $$0<x<\pi$$ (all option intervals lie here).

1. In the first quadrant $$0<x<\frac{\pi}{2},$$ both $$\sin x$$ and $$\cos x$$ are positive, so their product $$\sin x\cos x$$ is positive.

2. In the second quadrant $$\frac{\pi}{2}<x<\pi,$$ we have $$\sin x>0$$ but $$\cos x<0,$$ hence $$\sin x\cos x<0.$$

3. The remaining factor $$\sin^{2}x-\cos^{2}x$$ changes sign precisely when it is zero. Setting it to zero,

$$\sin^{2}x-\cos^{2}x=0\quad\Longrightarrow\quad\sin^{2}x=\cos^{2}x\quad\Longrightarrow\quad\tan^{2}x=1,$$

which holds at $$x=\frac{\pi}{4}$$ inside the first quadrant (and at $$x=\frac{3\pi}{4}$$ in the second).

Choose test points to locate where $$\sin^{2}x-\cos^{2}x$$ is positive or negative:

• For $$x=\frac{\pi}{6}$$ (which is <$$\frac{\pi}{4}$$), $$\sin^{2}x<\cos^{2}x,$$ so $$\sin^{2}x-\cos^{2}x<0.$$
• For $$x=\frac{\pi}{3}$$ (which is >$$\frac{\pi}{4}$$ but <$$\frac{\pi}{2}$$), $$\sin^{2}x>\cos^{2}x,$$ so $$\sin^{2}x-\cos^{2}x>0.$$

We now combine the signs.

First quadrant, $$0<x<\frac{\pi}{2}$$

• For $$0<x<\frac{\pi}{4}:$$ $$\sin x\cos x>0$$ and $$\sin^{2}x-\cos^{2}x<0,$$ so $$f'(x)<0.$$ Thus $$f(x)$$ decreases here.

• For $$\frac{\pi}{4}<x<\frac{\pi}{2}:$$ $$\sin x\cos x>0$$ and $$\sin^{2}x-\cos^{2}x>0,$$ so $$f'(x)>0.$$ Thus $$f(x)$$ increases here.

Second quadrant, $$\frac{\pi}{2}<x<\pi$$

Here $$\sin x\cos x<0.$$ Meanwhile $$\sin^{2}x-\cos^{2}x$$ stays positive until $$x=\frac{3\pi}{4}$$ and becomes negative afterwards, but the product with the negative $$\sin x\cos x$$ ensures $$f'(x)$$ remains negative throughout the whole second quadrant. Consequently, $$f(x)$$ decreases for all $$\frac{\pi}{2}<x<\pi.$$

Putting everything together, the derivative is positive (so $$f(x)$$ is increasing) only on

$$\left(\frac{\pi}{4},\;\frac{\pi}{2}\right).$$

This matches Option C exactly.

Hence, the correct answer is Option C.