The derivative of $ 2^{\frac{x}{\ln x}} $

Question

I am supposed to find the derivative of $ 2^{\frac{x}{\ln x}} $. My answer is $$ 2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot \frac{\ln x-x\cdot \frac{1}{x}}{\ln^{2}x}\cdot \frac{1}{x} .$$ Is it correct? Thanks.

Answer 1

You are almost correct. You have just an extra factor $1/x$ at the end. The correct derivative is $$D(2^{\frac{x}{\ln x}})=2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot D\left(\frac{x}{\ln x}\right)=2^{\frac{x}{\ln x}} \cdot \ln 2 \cdot \frac{\ln x-x\cdot \frac{1}{x}}{\ln^{2}x}.$$

Answer 2

Mathematica gives:

$$\frac{\log (2) 2^{\frac{x}{\log (x)}} (\log (x)-1)}{\log ^2(x)}$$

Answer 3

Let $y=2^\frac{x}{\ln{x}}$. Then

$$\ln{y}=\ln{2}^{\frac{x}{\ln{x}}}=\frac{x}{\ln{x}}\cdot\ln2$$

Now $$\frac{1}{y}\cdot\frac{dy}{dx}=\ln{2}\left[\frac{\ln{x}-1}{(\ln x)^2}\right]$$

Now multiply by $y$ and get

$$\frac{dy}{dx}=y\ln{2}\left[\frac{\ln{x}-1}{(\ln x)^2}\right]=2^\frac{x}{\ln{x}}\ln{2}\left[\frac{\ln{x}-1}{(\ln x)^2}\right]$$

You can do all sorts of algebra to reduce this but it is the derivative.