To solve this problem, we use the Leibniz rule for differentiating an integral with variable limits. The formula is:

$\frac{d}{dt} \int_{a(t)}^{b(t)} f(x) \, dx = f(b(t)) \cdot b'(t) - f(a(t)) \cdot a'(t)$

Here, the integral is:

$\frac{d}{dt} \int_1^{\tan t} \sin(x^2) \, dx$

1. Function $f(x) = \sin(x^2)$: This is the integrand.
2. Limits of integration: The lower limit is constant ($1$), so its derivative is $0$. The upper limit is $\tan t$, and its derivative is $\sec^2 t$.

Using the Leibniz rule:

$\frac{d}{dt} \int_1^{\tan t} \sin(x^2) \, dx = \sin((\tan t)^2) \cdot \frac{d}{dt}(\tan t) - \sin(1^2) \cdot \frac{d}{dt}(1)$

Since $\frac{d}{dt}(1) = 0$, the second term vanishes. The derivative becomes:

$\frac{d}{dt} \int_1^{\tan t} \sin(x^2) \, dx = \sin((\tan t)^2) \cdot \sec^2 t$

Final Answer:

$\frac{d}{dt} \int_1^{\tan t} \sin(x^2) \, dx = \sin((\tan t)^2) \cdot \sec^2 t$

Would you like further clarification or a step-by-step explanation of any part?

Related Questions:

1. How does the Leibniz rule work for integrals with variable limits?
2. What happens if both limits are variable?
3. Can this rule be extended to functions of multiple variables?
4. What are some real-world applications of derivatives of integrals?
5. How does the trigonometric identity for $\sec^2 t$ relate to this problem?

Tip:

Always double-check whether the limits of integration are variable or constant, as this determines how terms contribute to the derivative.