https://r2.erweima.ai/mathbot/image/1726678964991wzniexhx.jpg

It appears that the mathematical expression in your uploaded image is:

\[
(r^r + r^r)' + r^r \times r^r
\]

To solve this expression:

1. **Simplify the first term:** \(r^r + r^r = 2r^r\).
2. **Differentiate the first term:** Using the product rule for differentiation, the derivative of \(r^r\) can be computed as \(r^r(\log(r) + 1)\). Thus, \((2r^r)' = 2(r^r(\log(r) + 1)) = 2r^r\log(r) + 2r^r\).
3. **Simplify the second term:** \(r^r \times r^r = (r^r)^2 = r^{2r}\).

Thus, the expression simplifies to:

\[
2r^r\log(r) + 2r^r + r^{2r}
\]

Would you like further details on this derivation or do you have any other questions? Here are some potential follow-ups:

1. How is the derivative of \(r^r\) calculated?
2. What is the product rule for differentiation?
3. Can we explore more examples of differentiation with exponents?
4. How would this expression change if \(r\) is a specific number?
5. What are the graphical implications of these terms on a plot?

**Tip:** When dealing with expressions involving the same base raised to different powers, it’s often useful to simplify the expression first before applying any calculus operations. This can make the calculus work much clearer and simpler.

In this example, we simplify and solve the exponential expression (3^3 + 0.4)^1 + 3^3 × 3^3. The solution involves key mathematical concepts such as exponentiation, the power rule, and basic operations of addition and multiplication. Suitable for students in Grades 10-12.

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