if y=((x)^(logx))^log(logx) then find the value of dy/dx

To find the derivative of the function \( y = (x \log x) \log(\log x) \), we can apply the product rule and the chain rule of differentiation. This function consists of multiple components, so let’s break it down step by step.

Understanding the Components

The function can be viewed as the product of two parts: \( u = x \log x \) and \( v = \log(\log x) \). Therefore, we can express \( y \) as:

\( y = u \cdot v \)

Applying the Product Rule

According to the product rule, the derivative of a product \( uv \) is given by:

\( \frac{dy}{dx} = u'v + uv' \)

We need to compute \( u' \) and \( v' \) separately.

Finding \( u' \)

To differentiate \( u = x \log x \), we again use the product rule because it’s a product of \( x \) and \( \log x \):

Let:

• \( a = x \)
• \( b = \log x \)

Then:

\( u' = a'b + ab' \)

Here:

• \( a' = 1 \) (the derivative of \( x \))
• \( b' = \frac{1}{x} \) (the derivative of \( \log x \))

Plugging these into the product rule yields:

\( u' = (1)(\log x) + (x)\left(\frac{1}{x}\right) = \log x + 1 \)

Finding \( v' \)

Next, we differentiate \( v = \log(\log x) \). This requires the chain rule:

Let \( w = \log x \), so \( v = \log w \). The derivative is:

\( v' = \frac{1}{w} \cdot w' = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \)

Combining the Derivatives

Now we can substitute \( u, v, u', \) and \( v' \) back into the product rule:

\( \frac{dy}{dx} = (u'v) + (uv') \)

Substituting our values:

\( \frac{dy}{dx} = (\log x + 1)\log(\log x) + (x \log x)\left(\frac{1}{x \log x}\right) \)

This simplifies to:

\( \frac{dy}{dx} = (\log x + 1)\log(\log x) + 1 \)

Final Expression for the Derivative

The final expression for the derivative is:

\( \frac{dy}{dx} = (\log x + 1)\log(\log x) + 1 \)

This gives us a clear understanding of how the function \( y \) changes with respect to \( x \). Each component contributes to the overall rate of change, and by breaking it down using calculus rules, we can find the derivative effectively.