x2 + 4 , x > 0 dy dx

To find $\frac{dy}{dx}$ using logarithmic differentiation for the function:

$y = x^{x^2 + 4}$

we proceed as follows:

Step 1: Take the natural logarithm of both sides.

$\ln y = \ln(x^{x^2 + 4})$

Using the logarithmic property $\ln(a^b) = b \ln a$, we get:

$\ln y = (x^2 + 4) \ln x$

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

On the left-hand side, by implicit differentiation:

$\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$

On the right-hand side, use the product rule to differentiate $(x^2 + 4) \ln x$. Let $u = x^2 + 4$ and $v = \ln x$, then:

$\frac{d}{dx}\left((x^2 + 4)\ln x\right) = \frac{d}{dx}(u v) = \frac{du}{dx}v + u\frac{dv}{dx}$

Now, compute each derivative:

• $\frac{du}{dx} = 2x$
• $\frac{dv}{dx} = \frac{1}{x}$

So,

$\frac{d}{dx}\left((x^2 + 4)\ln x\right) = 2x \ln x + (x^2 + 4) \cdot \frac{1}{x}$

Simplify the second term:

$\frac{d}{dx}\left((x^2 + 4)\ln x\right) = 2x \ln x + x + \frac{4}{x}$

Step 3: Substitute into the implicit differentiation equation.

Now, we have:

$\frac{1}{y} \frac{dy}{dx} = 2x \ln x + x + \frac{4}{x}$

Step 4: Solve for $\frac{dy}{dx}$.

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

$\frac{dy}{dx} = y\left(2x \ln x + x + \frac{4}{x}\right)$

Step 5: Substitute $y = x^{x^2 + 4}$ back into the equation.

$\frac{dy}{dx} = x^{x^2 + 4}\left(2x \ln x + x + \frac{4}{x}\right)$

This is the derivative of the given function.

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Let me know if you'd like further clarification or more details.

Related Questions:

1. How can logarithmic differentiation simplify complex exponentials?
2. What is the product rule, and how does it apply in this case?
3. Can logarithmic differentiation be applied to non-exponential functions?
4. What are the benefits of implicit differentiation in finding derivatives?
5. How does the chain rule play a role in logarithmic differentiation?

Tip:

For complex exponentials like $x^{x^2 + 4}$, logarithmic differentiation often simplifies the process, avoiding messy power rule applications.