2023-03-10

How to differentiate $y={\left(\mathrm{ln}x\right)}^{\mathrm{ln}x}$?

Karbamidjts

=$\frac{dy}{dx}=d\frac{{\left(\mathrm{ln}x\right)}^{\mathrm{ln}x}}{dx}$
=$\frac{dy}{dx}=\left(\mathrm{ln}x\right)\cdot {\left(\mathrm{ln}x\right)}^{\left(\mathrm{ln}x\right)-1}\cdot \frac{1}{x}$
formulae used $n{x}^{n-1}$=$d\frac{{x}^{n}}{dx}$
concept used MOSTLY importantly chain rule,and formula

unieventos8l9

To begin, we must convert this function into an equivalent function in order to make it more manageable with more familiar rules. Therefore, we turn ${\left(\mathrm{ln}x\right)}^{\mathrm{ln}x}$ into ${e}^{\left(\mathrm{ln}\left(\mathrm{ln}x\right)\right)\left(\mathrm{ln}x\right)}$. This works because, due to raising e to any power is an inverse function to taking the natural log of any number, ${e}^{\mathrm{ln}\left(x\right)}=x$ for any x. This may seem like a more complicated function we have made to take the derivative, but trust me, it makes thins simpler. Thus, since we know that the derivative of ${e}^{u}$ is $u\prime \cdot {e}^{u}$ we'll take the derivative of ${e}^{\left(\mathrm{ln}\left(\mathrm{ln}x\right)\right)\left(\mathrm{ln}x\right)}$ using this rule.
This means we must take the derivative of the exponent and multiply by the original function. This brings us a whole new, but much simpler problem. Now we must find the derivative of $\mathrm{ln}\left(\mathrm{ln}x\right)\cdot \mathrm{ln}x$ using the product rule which means that the derivative of any two functions multiplied is the first function's derivative multiplied by the second function plus the first function multiplied by the second function's derivative.
Side note: since the derivative of ln(u) is $\frac{u\prime }{u}$ the derivative of ln(lnx) is $\frac{1}{\mathrm{ln}x}$ multiplied by the derivative of ln x which is $\frac{1}{x}$. This means the derivative of ln (ln x) is $\frac{1}{x\cdot \mathrm{ln}x}$.
This gives us the derivative of $\mathrm{ln}\left(\mathrm{ln}x\right)\cdot \mathrm{ln}x$ which is $\frac{\mathrm{ln}x}{x\cdot \mathrm{ln}x}+\frac{\mathrm{ln}\left(\mathrm{ln}x\right)}{x}$. If we do some cancellation we get: $\frac{1}{x}+\frac{\mathrm{ln}\left(\mathrm{ln}x\right)}{x}$, but since they both have denominators of x we can combine them to get $\frac{\mathrm{ln}\left(\mathrm{ln}x\right)+1}{x}$. This is the original exponent's derivative, which we will multiply with the original function.
This gives us: $\left(\frac{\mathrm{ln}\left(\mathrm{ln}x\right)+1}{x}\right){e}^{\mathrm{ln}\left(\mathrm{ln}x\right)\cdot \mathrm{ln}x}$ which can also be written as $\left(\frac{\mathrm{ln}\left(\mathrm{ln}x\right)+1}{x}\right){\left(\mathrm{ln}x\right)}^{\mathrm{ln}x}$ and THIS is the final answer to the derivative of ${\left(\mathrm{ln}x\right)}^{\mathrm{ln}x}$.

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