To differentiate the given function:

$y = \left(10x + 1\right)^7 - \log_5\left(x^2 + e^5\right),$

we proceed step by step.

Step 1: Differentiate $(10x + 1)^7$

We apply the chain rule:
$\frac{d}{dx} \left( f(g(x)) \right) = f'(g(x)) \cdot g'(x).$ Here, $f(u) = u^7$ and $g(x) = 10x + 1$.

• $f'(u) = 7u^6$
• $g'(x) = 10$.

Thus,
$\frac{d}{dx} \left( (10x + 1)^7 \right) = 7(10x + 1)^6 \cdot 10 = 70(10x + 1)^6.$

Step 2: Differentiate $\log_5(x^2 + e^5)$

To differentiate a logarithm with a base $a$, we use:
$\frac{d}{dx} \left( \log_a(u) \right) = \frac{1}{u \ln(a)} \cdot \frac{du}{dx}.$ Here, $u = x^2 + e^5$ and $a = 5$.

1. First, $\frac{du}{dx} = \frac{d}{dx}(x^2 + e^5) = 2x$ since $e^5$ is a constant.
2. The derivative of $\log_5(u)$ is:
   $\frac{d}{dx} \left( \log_5(x^2 + e^5) \right) = \frac{1}{(x^2 + e^5) \ln(5)} \cdot 2x.$

Simplify:
$\frac{d}{dx} \left( \log_5(x^2 + e^5) \right) = \frac{2x}{(x^2 + e^5) \ln(5)}.$

Step 3: Combine Results

The derivative of $y$ is:
$y' = 70(10x + 1)^6 - \frac{2x}{(x^2 + e^5) \ln(5)}.$

Final Answer:

$\boxed{y' = 70(10x + 1)^6 - \frac{2x}{(x^2 + e^5) \ln(5)}}$

Would you like me to verify or expand on any part of the solution?

Related Questions:

1. What is the derivative of $(3x + 2)^5$?
2. How do you differentiate $\log_a(u)$ for general $a$ and $u$?
3. What is the derivative of $\ln(x^2 + 1)$?
4. How do you handle higher-order derivatives of a chain-rule function?
5. How does the choice of logarithmic base affect differentiation?

Tip:

To handle logarithms in differentiation, converting to the natural logarithm form $\ln(x)$ often simplifies calculations. For example:
$\log_a(x) = \frac{\ln(x)}{\ln(a)}.$