https://file.aiquickdraw.com/mathbot/file/172672229621245qmtgrz.jpg

The question shown in the image asks to compute \(\frac{dy}{dx}\) given that \(x = t^2\) and \(y = t^3\).

We can solve this using the **chain rule**.

### Steps:
1. We know \(x = t^2\) and \(y = t^3\).
2. We are looking for \(\frac{dy}{dx}\).

By the chain rule:

\[
\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}
\]

Let's compute each part:

- \(\frac{dy}{dt} = \frac{d}{dt}(t^3) = 3t^2\)
- \(\frac{dx}{dt} = \frac{d}{dt}(t^2) = 2t\)

Now, since \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2t}\), we can substitute this back into the chain rule:

\[
\frac{dy}{dx} = 3t^2 \cdot \frac{1}{2t} = \frac{3t}{2}
\]

Thus, the value of \(\frac{dy}{dx}\) is \(\frac{3t}{2}\), which corresponds to option (A) in the image.

Would you like further details or clarification?

Here are 5 related questions:
1. How is the chain rule applied in differentiation of composite functions?
2. What happens when the variables in \(x\) and \(y\) are not functions of \(t\)?
3. Can you apply implicit differentiation when no explicit \(t\)-dependency is given?
4. How would you differentiate a parametric equation if more than two variables were involved?
5. What is the geometrical interpretation of \(\frac{dy}{dx}\) in parametric equations?

**Tip:** When using the chain rule for parametric equations, always differentiate both \(x(t)\) and \(y(t)\) with respect to the parameter \(t\) first, and then combine using the chain rule.

If x = t^2 and y = t^3, then dy/dx is equal to:

Learn how to compute dy/dx for the parametric equations x = t^2 and y = t^3 using the chain rule. Step-by-step solution explaining the application of the chain rule and its components in differentiation. Suitable for students in Grades 10-12.

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