1. Given function

$$f(x)=\sin^{-1}\left(\frac{2\cdot 3^x}{1+9^x}\right)$$

We need to find:

$$f'\left(-\frac12\right)$$

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2. Simplify the expression inside

Since

$$9^x=(3^2)^x=3^{2x},$$

let

$$t=3^x.$$

Then the argument of $\sin^{-1}$ becomes

$$\frac{2t}{1+t^2}.$$

Now use the identity:

$$\sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta}.$$

If we take

$$t=\tan\theta,$$

then

$$\frac{2t}{1+t^2}=\sin 2\theta.$$

So with $t=3^x$, we can write

$$\frac{2\cdot 3^x}{1+3^{2x}}=\sin\big(2\tan^{-1}(3^x)\big).$$

Hence

$$f(x)=\sin^{-1}\left(\sin\big(2\tan^{-1}(3^x)\big)\right).$$

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3. Check principal value near $x=-\frac12$

At

$$x=-\frac12,$$

we have

$$3^x=3^{-1/2}=\frac{1}{\sqrt3}.$$

Thus

$$2\tan^{-1}\left(\frac1{\sqrt3}\right)=2\cdot \frac{\pi}{6}=\frac{\pi}{3}.$$

Since $\frac{\pi}{3}\in \left[-\frac\pi2,\frac\pi2\right]$, the principal value of $\sin^{-1}(\sin u)$ equals $u$ here. Therefore, in a neighborhood of $x=-\frac12$,

$$f(x)=2\tan^{-1}(3^x).$$

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4. Differentiate

$$f'(x)=2\cdot \frac{1}{1+(3^x)^2}\cdot \frac{d}{dx}(3^x)$$

and

$$\frac{d}{dx}(3^x)=3^x\ln 3.$$

So,

$$f'(x)=\frac{2\cdot 3^x\ln 3}{1+3^{2x}}.$$

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5. Evaluate at $x=-\frac12$

Since

$$3^{-1/2}=\frac1{\sqrt3}, \qquad 3^{2(-1/2)}=3^{-1}=\frac13,$$

we get

$$f'\left(-\frac12\right)=\frac{2\cdot \frac1{\sqrt3}\ln 3}{1+\frac13}.$$

Now,

$$1+\frac13=\frac43,$$

so

$$f'\left(-\frac12\right)=\frac{2\ln 3}{\sqrt3}\cdot \frac34 =\frac{3\ln 3}{2\sqrt3}.$$

Rationalizing/simplifying:

$$\frac{3}{2\sqrt3}=\frac{\sqrt3}{2}.$$

Hence

$$f'\left(-\frac12\right)=\frac{\sqrt3}{2}\ln 3.$$

Also,

$$\ln \sqrt3=\frac12\ln 3,$$

therefore

$$\frac{\sqrt3}{2}\ln 3=\sqrt3\ln \sqrt3.$$

So,

$$f'\left(-\frac12\right)=\sqrt3\log_e \sqrt3.$$

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6. Compare with options

• A: $-\sqrt3\log_e\sqrt3$
• B: $\sqrt3\log_e\sqrt3$
• C: $-\sqrt3\log_e 3$
• D: $\sqrt3\log_e 3$

Thus the correct option is:

$$\boxed{\text{B}}$$