find first derivative $$ f\left(t\right)=\left(t^2+1\right)^5 $$

To find the first derivative of the function:

\[
f(t) = (t^2 + 1)^5
\]

we use the **chain rule**, which states:

\[
\frac{d}{dt} [g(h(t))] = g'(h(t)) \cdot h'(t).
\]

### Step 1: Define Inner and Outer Functions
Let:
- $ h(t) = t^2 + 1 $
- $ g(x) = x^5 $, where $ x = h(t) $

### Step 2: Differentiate Outer Function
\[
\frac{d}{dx} [x^5] = 5x^4
\]

Substituting $ x = t^2 + 1 $:

\[
\frac{d}{dt} [(t^2 + 1)^5] = 5 (t^2 + 1)^4 \cdot \frac{d}{dt} [t^2 + 1].
\]

### Step 3: Differentiate Inner Function
\[
\frac{d}{dt} [t^2 + 1] = 2t.
\]

### Step 4: Multiply Both Derivatives
\[
f'(t) = 5 (t^2 + 1)^4 \cdot 2t.
\]

### Final Answer:
\[
f'(t) = 10t (t^2 + 1)^4.
\]

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

Here are some related questions:
1. What is the second derivative of $ f(t) $?
2. How does the chain rule apply in this problem?
3. What happens if we change the exponent to 6 instead of 5?
4. Can we use logarithmic differentiation here?
5. How do we find the critical points of $ f(t) $?

**Tip:** The chain rule is essential when differentiating composite functions—always identify the inner and outer functions first!

Learn how to find the first derivative of f(t) = (t^2 + 1)^5 using the chain rule. This detailed step-by-step solution demonstrates identifying inner and outer functions, applying differentiation rules, and simplifying the result to obtain f'(t) = 10t (t^2 + 1)^4. Suitable for Grades 11-12 and college-level calculus students.

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