If y = (tan x)ˢᶦⁿ ˣ, then dy/dx is equal to

The correct option is "None of these".

To find the derivative of the given function, we use the method of logarithmic differentiation.

Let the given function be:
$$y={(tanx)}^{sinx}$$

Taking the natural logarithm (log to the base e) on both sides:
$$logy=log\left[{(tanx)}^{sinx}\right]$$

Using the logarithmic property $log\left(a^b\right)=bloga$, we can rewrite the equation as:
$$logy=sinx\cdot log(tanx)$$

Now, differentiating both sides with respect to $x$:
$$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[sinx\cdot log(tanx)]$$

Applying the product rule of differentiation, $\frac{d}{dx}\left[uv\right]=u\frac{dv}{dx}+v\frac{du}{dx}$:
$$\frac{1}{y}\frac{dy}{dx}=sinx\cdot \frac{d}{dx}[log(tanx)]+log(tanx)\cdot \frac{d}{dx}(sinx)$$

Using the chain rule, the derivative of $log(tanx)$ is $\frac{1}{tanx}\cdot {sec}^{2}x$:
$$\frac{1}{y}\frac{dy}{dx}=sinx\cdot [\frac{1}{tanx}\cdot {sec}^{2}x]+log(tanx)\cdot cosx$$

Let us simplify the first term:
$$sinx\cdot \frac{cosx}{sinx}\cdot \frac{1}{{cos}^{2}x}=\frac{1}{cosx}=secx$$

Substituting this back into our expression:
$$\frac{1}{y}\frac{dy}{dx}=secx+cosx\cdot log(tanx)$$

Multiplying both sides by $y$:
$$\frac{dy}{dx}=y[secx+cosx\cdot log(tanx)]$$

Replacing $y$ with its original value ${(tanx)}^{sinx}$:
$$\frac{dy}{dx}={(tanx)}^{sinx}[secx+cosx\cdot log(tanx)]$$

Comparing this derivative with the options:
Option 1: $secx+cosx$ (Incorrect)
Option 2: $secx+logtanx$ (Incorrect)
Option 3: ${(tanx)}^{sinx}$ (Incorrect)

Since the derived expression does not match any of the first three options, the correct answer is "None of these".