find the indicated derivative
dy/dt where y=(sin(t)tan(t^2+1)

To find the derivative \(\frac{dy}{dt}\) where \( y = \sin(t) \cdot \tan(t^2 + 1) \), we will apply the **product rule** and the **chain rule**.

The product rule states:

\[
\frac{d}{dt}[u(t) \cdot v(t)] = u'(t) \cdot v(t) + u(t) \cdot v'(t)
\]

Here, let:

\[
u(t) = \sin(t) \quad \text{and} \quad v(t) = \tan(t^2 + 1)
\]

Now, we'll compute the derivatives of both \( u(t) \) and \( v(t) \).

### Step 1: Derivative of \( u(t) = \sin(t) \)
\[
u'(t) = \cos(t)
\]

### Step 2: Derivative of \( v(t) = \tan(t^2 + 1) \)

To differentiate \( v(t) = \tan(t^2 + 1) \), we need to apply the chain rule. The derivative of \( \tan(x) \) is \( \sec^2(x) \), so:

\[
v'(t) = \frac{d}{dt} \left[\tan(t^2 + 1)\right] = \sec^2(t^2 + 1) \cdot \frac{d}{dt}(t^2 + 1)
\]

Now, compute \( \frac{d}{dt}(t^2 + 1) \):

\[
\frac{d}{dt}(t^2 + 1) = 2t
\]

Thus,

\[
v'(t) = \sec^2(t^2 + 1) \cdot 2t
\]

### Step 3: Apply the Product Rule

Now, we can use the product rule:

\[
\frac{dy}{dt} = u'(t) \cdot v(t) + u(t) \cdot v'(t)
\]

Substitute the values:

\[
\frac{dy}{dt} = \cos(t) \cdot \tan(t^2 + 1) + \sin(t) \cdot \sec^2(t^2 + 1) \cdot 2t
\]

This is the derivative \(\frac{dy}{dt}\).

Would you like further clarification or details on any part of the solution?

Here are 5 related questions that expand on the concept:
1. How would you apply the product rule to three functions?
2. What is the chain rule and why is it essential in this problem?
3. How would you find the second derivative of \( y \)?
4. How does the derivative of a trigonometric function differ from its integral?
5. Can you explain when to apply implicit differentiation in calculus?

**Tip:** Always simplify trigonometric identities before differentiating to make your computations easier!

find the indicated derivative dy/dt where y = (sin(t)tan(t^2+1))

Learn how to find the derivative dy/dt where y = sin(t) * tan(t^2 + 1). This calculus problem involves using the product rule and chain rule to differentiate trigonometric functions. Suitable for high school students in Grades 11-12, this solution provides a step-by-step approach to understanding derivatives of complex functions.

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