We are tasked with finding the derivative of the function $y = x^{4x + 3}$ using logarithmic differentiation.

Let's go step by step:

Step 1: Apply logarithmic differentiation

Start by taking the natural logarithm (ln) of both sides of the equation:

$\ln y = \ln \left( x^{4x + 3} \right)$

Using the logarithmic property $\ln(a^b) = b \ln(a)$, we simplify the right-hand side:

$\ln y = (4x + 3) \ln x$

Step 2: Differentiate both sides with respect to $x$

Now, differentiate both sides with respect to $x$. Remember to apply the product rule to the right-hand side:

$\frac{d}{dx} \left( \ln y \right) = \frac{d}{dx} \left( (4x + 3) \ln x \right)$

Using the chain rule on the left-hand side and the product rule on the right-hand side:

$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(4x + 3) \cdot \ln x + (4x + 3) \cdot \frac{d}{dx}(\ln x)$

Now, compute the derivatives on the right-hand side: $\frac{d}{dx}(4x + 3) = 4 \quad \text{and} \quad \frac{d}{dx}(\ln x) = \frac{1}{x}$

Thus, we have:

$\frac{1}{y} \cdot \frac{dy}{dx} = 4 \ln x + \frac{4x + 3}{x}$

Step 3: Solve for $\frac{dy}{dx}$

Multiply both sides by $y$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \left( 4 \ln x + \frac{4x + 3}{x} \right)$

Finally, substitute $y = x^{4x + 3}$ back into the equation:

$\frac{dy}{dx} = x^{4x + 3} \left( 4 \ln x + \frac{4x + 3}{x} \right)$

Final Result:

$\frac{dy}{dx} = x^{4x + 3} \left( 4 \ln x + \frac{4x + 3}{x} \right)$

This is the derivative of $y$ with respect to $x$ using logarithmic differentiation.

Would you like further clarification or details on any of the steps?

Here are 5 related questions to expand your understanding:

1. How is logarithmic differentiation different from standard differentiation?
2. Why is the product rule necessary when differentiating $(4x + 3) \ln x$?
3. Can logarithmic differentiation be applied to functions involving sums of powers?
4. What other types of functions benefit from logarithmic differentiation?
5. How would you differentiate $y = x^{x^2}$ using logarithmic differentiation?

Tip: Logarithmic differentiation is particularly useful when dealing with functions where the variable appears both in the base and the exponent!