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$y = \ln(x) + 6x, \quad \text{Find} \ \frac{dy}{dx} + y.$

The problem asks for the derivative of $y$ with respect to $x$ and then adds the original function $y$ to this derivative.

Step-by-Step Solution:

1. Find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} \left(\ln(x) + 6x\right)$ The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$. The derivative of $6x$ with respect to $x$ is $6$. $\frac{dy}{dx} = \frac{1}{x} + 6$

2. Add $y$ to $\frac{dy}{dx}$: $\frac{dy}{dx} + y = \left(\frac{1}{x} + 6\right) + \left(\ln(x) + 6x\right)$ Simplify the expression: $\frac{dy}{dx} + y = \frac{1}{x} + 6 + \ln(x) + 6x$

3. Final Answer: $\frac{dy}{dx} + y = \frac{1}{x} + \ln(x) + 6x + 6$

So, the correct answer corresponds to the option that represents this expression.

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Related Questions:

1. How do you differentiate logarithmic functions in general?
2. What are the basic rules of differentiation?
3. How do you find the derivative of a sum of functions?
4. What are some practical applications of derivatives in real life?
5. How does the derivative of a product of functions differ from the sum of functions?

Tip: When finding the derivative of a logarithmic function, always remember that the derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$.