To find the derivative of $y = 4x^x$ using logarithmic differentiation, follow these steps:

Step 1: Take the natural logarithm of both sides

$\ln(y) = \ln(4x^x)$

Using the logarithmic property $\ln(ab) = \ln(a) + \ln(b)$: $\ln(y) = \ln(4) + \ln(x^x)$

Now use the property $\ln(x^n) = n\ln(x)$: $\ln(y) = \ln(4) + x\ln(x)$

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

We use implicit differentiation here. The derivative of $\ln(y)$ is $\frac{1}{y} \cdot \frac{dy}{dx}$, and we differentiate the right-hand side term by term: $\frac{1}{y} \cdot \frac{dy}{dx} = 0 + \frac{d}{dx}[x\ln(x)]$

Differentiate $x\ln(x)$ using the product rule: $\frac{d}{dx}[x\ln(x)] = \ln(x) + 1$

So: $\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + 1$

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

Multiply through by $y$: $\frac{dy}{dx} = y (\ln(x) + 1)$

Substitute $y = 4x^x$ back into the equation: $\frac{dy}{dx} = 4x^x (\ln(x) + 1)$

Final Answer:

$\boxed{\frac{dy}{dx} = 4x^x (\ln(x) + 1)}$

Let me know if you’d like a more detailed breakdown of any step or if you have further questions!

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5 Related Questions:

1. How is logarithmic differentiation helpful for solving $y = x^x$?
2. Can logarithmic differentiation be used for $y = x^{\sin(x)}$? If so, how?
3. What is the derivative of $y = x^x$ (similar problem without the constant factor)?
4. How does the product rule apply to logarithmic differentiation?
5. What are the limitations of logarithmic differentiation?

Tip:

Logarithmic differentiation is especially useful when the variable appears both in the base and the exponent of a function!