To solve this, I will extract the function from the image and compute the derivative.

The function appears to be:

$y = \sqrt{3} \cdot \sqrt{x^2 - 1}$

Let’s differentiate this step by step.

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Step 1: Rewrite the function

The given function can be rewritten as:

$y = \sqrt{3} \cdot (x^2 - 1)^{1/2}$

Step 2: Differentiate using the chain rule

The derivative of $y$ with respect to $x$ is:

$\frac{dy}{dx} = \sqrt{3} \cdot \frac{d}{dx} \left( (x^2 - 1)^{1/2} \right)$

Now, apply the chain rule to differentiate $(x^2 - 1)^{1/2}$:

$\frac{d}{dx} \left( (x^2 - 1)^{1/2} \right) = \frac{1}{2} (x^2 - 1)^{-1/2} \cdot \frac{d}{dx}(x^2 - 1)$

The derivative of $x^2 - 1$ is $2x$. So:

$\frac{d}{dx} \left( (x^2 - 1)^{1/2} \right) = \frac{1}{2} (x^2 - 1)^{-1/2} \cdot 2x$

The $2$ cancels out:

$\frac{d}{dx} \left( (x^2 - 1)^{1/2} \right) = \frac{x}{\sqrt{x^2 - 1}}$

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Step 3: Substitute back

Now substitute this into the derivative of $y$:

$\frac{dy}{dx} = \sqrt{3} \cdot \frac{x}{\sqrt{x^2 - 1}}$

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Final Answer:

$\frac{dy}{dx} = \frac{\sqrt{3} \cdot x}{\sqrt{x^2 - 1}}$

Would you like further clarification or additional steps? 😊

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5 Related Questions:

1. How do we handle derivatives of square roots in general?
2. What happens if $x^2 - 1 \leq 0$? What are the restrictions on $x$?
3. How do we apply the product rule if there are two terms in a function like this?
4. Can we simplify the derivative further if given specific values of $x$?
5. How do derivatives involving radicals appear in applications, like physics?

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Tip:

Always check the domain of the original function when working with square roots—it ensures the function is well-defined! For this case, $x^2 - 1 > 0$, so $x > 1$ or $x < -1$.