The image shows the integral expression:

$y = \int_{\sqrt{x}}^{\frac{\pi}{6}} \theta \tan(\theta) \, d\theta$

This is an application of the Fundamental Theorem of Calculus, and we are asked to find its derivative with respect to $x$.

Step-by-step solution:

To apply the Fundamental Theorem of Calculus (Part 1), we first recognize that the function $y$ is defined as an integral with a variable limit of integration ($\sqrt{x}$).

The derivative of such an integral with respect to $x$ can be found using Leibniz’s rule:

$\frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(\theta) \, d\theta \right) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)$

Here, $a(x) = \sqrt{x}$, and $b(x) = \frac{\pi}{6}$. Also, $f(\theta) = \theta \tan(\theta)$.

• The derivative of the upper limit $b(x) = \frac{\pi}{6}$ is 0 since it is a constant.
• The derivative of the lower limit $a(x) = \sqrt{x}$ is $a'(x) = \frac{1}{2\sqrt{x}}$.

Now applying this to the expression:

$\frac{dy}{dx} = \left( 0 \cdot f\left(\frac{\pi}{6}\right) \right) - \left( \theta \tan(\theta) \bigg|_{\theta = \sqrt{x}} \right) \cdot \frac{1}{2\sqrt{x}}$

Thus, the derivative is:

$\frac{dy}{dx} = - \left( \sqrt{x} \cdot \tan(\sqrt{x}) \right) \cdot \frac{1}{2\sqrt{x}}$

Simplifying:

$\frac{dy}{dx} = - \frac{\sqrt{x} \cdot \tan(\sqrt{x})}{2\sqrt{x}} = - \frac{\tan(\sqrt{x})}{2}$

Final answer:

$\frac{dy}{dx} = - \frac{\tan(\sqrt{x})}{2}$

Would you like more details on any specific part of the solution?

Here are some related questions to further explore the topic:

1. What is the Leibniz rule for differentiating integrals with variable limits?
2. How does the Fundamental Theorem of Calculus Part 1 apply to derivatives?
3. Can you compute the integral of $\theta \tan(\theta)$?
4. How do you handle constant limits of integration in the Fundamental Theorem of Calculus?
5. What is the role of trigonometric functions like $\tan(\theta)$ in calculus?

Tip: When differentiating integrals with variable limits, always apply Leibniz’s rule, which accounts for the behavior of the integrand and the variable limits.