To differentiate the vector function $\mathbf{r}(t) = e^{10t} \mathbf{i} + e^{5t} \mathbf{j}$, we take the derivative of each component separately with respect to $t$.

Step 1: Differentiate the $\mathbf{i}$-component

The $\mathbf{i}$-component is $e^{10t}$, and the derivative of $e^{10t}$ with respect to $t$ is: $\frac{d}{dt} \left( e^{10t} \right) = 10e^{10t}$

Step 2: Differentiate the $\mathbf{j}$-component

The $\mathbf{j}$-component is $e^{5t}$, and the derivative of $e^{5t}$ with respect to $t$ is: $\frac{d}{dt} \left( e^{5t} \right) = 5e^{5t}$

Step 3: Combine the results

Now, we combine the differentiated components: $\mathbf{r}'(t) = 10e^{10t} \mathbf{i} + 5e^{5t} \mathbf{j}$

Thus, the derivative of the vector function is: $\boxed{\mathbf{r}'(t) = 10e^{10t} \mathbf{i} + 5e^{5t} \mathbf{j}}$

Would you like more details on vector differentiation or any other questions about this process? Here are some related questions to explore further:

1. How do we compute higher-order derivatives of vector functions?
2. What is the significance of the vector derivative in physics?
3. How would we compute the magnitude of $\mathbf{r}'(t)$?
4. Can we use this derivative to find the tangent vector to a curve?
5. How does this concept apply in fields like engineering or computer graphics?

Tip: When differentiating exponentials, remember that the derivative of $e^{kt}$ with respect to $t$ is always $ke^{kt}$.