Let f(t) = 5t^4 - 6t + 2e^t then f' (t)= 

To find the derivative \( f'(t) \) of the function \( f(t) = 5t^4 - 6t + 2e^t \), we will differentiate each term separately.

1. **First term**: \( 5t^4 \)
   - Using the power rule \( \frac{d}{dt}[t^n] = n t^{n-1} \), we get:
   \[
   \frac{d}{dt}[5t^4] = 4 \cdot 5t^{3} = 20t^3
   \]

2. **Second term**: \( -6t \)
   - Using the rule \( \frac{d}{dt}[t] = 1 \), we get:
   \[
   \frac{d}{dt}[-6t] = -6
   \]

3. **Third term**: \( 2e^t \)
   - The derivative of \( e^t \) is simply \( e^t \), so:
   \[
   \frac{d}{dt}[2e^t] = 2e^t
   \]

Now, summing these derivatives together:

\[
f'(t) = 20t^3 - 6 + 2e^t
\]

Thus, the derivative of \( f(t) \) is:

\[
f'(t) = 20t^3 - 6 + 2e^t
\]

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. What is the second derivative \( f''(t) \) of \( f(t) = 5t^4 - 6t + 2e^t \)?
2. How would you find the critical points of \( f(t) = 5t^4 - 6t + 2e^t \)?
3. What is the behavior of \( f(t) \) as \( t \to \infty \)?
4. Can you apply the chain rule to find the derivative of a composite function involving \( f(t) \)?
5. What are the applications of derivatives in real-world problems?

**Tip**: When finding derivatives, always check if each term involves a product, quotient, or composite function, as these require additional rules (product rule, quotient rule, or chain rule).

Discover how to compute the derivative of f(t) = 5t^4 - 6t + 2e^t using the power rule and the derivative of exponential functions. This problem involves key calculus concepts and is suitable for Grades 11-12.

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