We are tasked with differentiating the function:

$y = (\log{x})^{\cos{(x^3 - 3x)}}, \quad \log{x} > 0$

To find $\frac{dy}{dx}$, we will apply logarithmic differentiation. Here's the step-by-step process:

Step 1: Apply the natural logarithm to both sides

Taking the natural logarithm on both sides will simplify the exponentiation:

$\ln{y} = \ln{\left((\log{x})^{\cos{(x^3 - 3x)}}\right)}$

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

$\ln{y} = \cos{(x^3 - 3x)} \cdot \ln{(\log{x})}$

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

Now we differentiate implicitly with respect to $x$. Start by differentiating the left-hand side:

$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \cos{(x^3 - 3x)} \ln{(\log{x})} \right)$

We will use the product rule on the right-hand side. Let’s denote:

• $u = \cos{(x^3 - 3x)}$
• $v = \ln{(\log{x})}$

The product rule states that $\frac{d}{dx}(uv) = u'v + uv'$.

Step 2.1: Differentiate $u = \cos{(x^3 - 3x)}$

Using the chain rule:

$u' = -\sin{(x^3 - 3x)} \cdot \frac{d}{dx}(x^3 - 3x) = -\sin{(x^3 - 3x)} \cdot (3x^2 - 3)$

Step 2.2: Differentiate $v = \ln{(\log{x})}$

Using the chain rule again:

$v' = \frac{1}{\log{x}} \cdot \frac{d}{dx}(\log{x}) = \frac{1}{\log{x}} \cdot \frac{1}{x} = \frac{1}{x \log{x}}$

Step 3: Combine using the product rule

Now, applying the product rule:

$\frac{d}{dx} \left( \cos{(x^3 - 3x)} \ln{(\log{x})} \right) = u'v + uv'$

Substitute the values of $u', u, v,$ and $v'$:

$\frac{d}{dx} \left( \cos{(x^3 - 3x)} \ln{(\log{x})} \right) = -\sin{(x^3 - 3x)}(3x^2 - 3) \ln{(\log{x})} + \cos{(x^3 - 3x)} \cdot \frac{1}{x \log{x}}$

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

Now, we have:

$\frac{1}{y} \frac{dy}{dx} = -\sin{(x^3 - 3x)}(3x^2 - 3) \ln{(\log{x})} + \cos{(x^3 - 3x)} \cdot \frac{1}{x \log{x}}$

Finally, multiply both sides by $y$ to isolate $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \left[ -\sin{(x^3 - 3x)}(3x^2 - 3) \ln{(\log{x})} + \cos{(x^3 - 3x)} \cdot \frac{1}{x \log{x}} \right]$

Since $y = (\log{x})^{\cos{(x^3 - 3x)}}$, we substitute this back:

$\frac{dy}{dx} = (\log{x})^{\cos{(x^3 - 3x)}} \left[ -\sin{(x^3 - 3x)}(3x^2 - 3) \ln{(\log{x})} + \cos{(x^3 - 3x)} \cdot \frac{1}{x \log{x}} \right]$

This is the derivative of the given function.

Would you like further details on any step or have any questions?

Related Questions:

1. How does logarithmic differentiation simplify exponentials?
2. Why is the chain rule necessary in this differentiation?
3. Can this function have critical points? If so, how can they be found?
4. How does the product rule apply to differentiating more complex functions?
5. How can we use the derivative to analyze the behavior of the function?

Tip:

When dealing with complex products or powers involving logarithms, logarithmic differentiation can be an efficient technique to simplify the process.