Wolfram Alpha is doing something I don't understand, please help

Question

I am using Newton Raphson method to obtain the velocity of a chemical reaction, and I needed to derive the next equation:$$\frac{d}{dk}\left(\left(\frac{0.35}{k}\right)^{\frac{k}{k-0.35}}\right)= -0.3285178\left(\frac{1}{k}\right)^{1.65}$$ And the answer they give me is correct: $-0.3285178\left(\frac{1}{k}\right)^{1.65}$The result in the method is $0.175$ but I don't understand how did wolfram alpha got that answer. They go from the original equation to another one which I can not understand $$\frac{d}{dk}\left(\left(\frac{0.35}{k}\right)^{\frac{k}{k-0.35}}\right) \to \frac{d}{dk}\left(0.50541\left(\frac{1}{k}\right)^{0.65}\right) = -0.3285178\left(\frac{1}{k}\right)^{1.65}$$I would appreciate if someone can explain to me how does the first convert to the second. I'll let here the link to the problem https://www.wolframalpha.com/input/?i=d%2Fdk+%280.35%2Fk%29%5E%28k%2Fk-0.35%29 Thanks

Answer 1

You didn't use parentheses. So first $k/k$ is evaluated i.e., $1$, then $k/k - 0.35 = 1 - 0.35 = 0.65$

Then, $0.35^{0.65} = 0.50541$

So use $(k/\color{red}(k-0.35\color{red}))$ to get the derivative you require.

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Let $$y = \left(\frac{0.35}k\right)^{\frac k{k-0.35}} = e^{\ln\left(\left(\frac{0.35}k\right)^{\frac k{k-0.35}}\right)} = e^{\frac k{k-0.35}\ln(\frac{0.35}k)}$$

$$y' =e^{\frac k{k-0.35}\ln(\frac{0.35}k)}\left[\frac{k}{k-0.35}\cdot\frac{k}{0.35}\cdot\frac{-0.35}{k^2}+ \ln\left(\frac{0.35}{k}\right)\left(\frac{1}{k-0.35}- \frac{k}{(k-0.35)^2}\right)\right ] $$

$$y' =\left(\frac{0.35}k\right)^{\frac k{k-0.35}} \left[\frac{-1}{k-0.35}+ \ln\left(\frac{0.35}{k}\right)\left(\frac{1}{k-0.35}- \frac{k}{(k-0.35)^2}\right)\right ]$$

or

$$y' = -\frac{\left(\frac{0.35}k\right)^{\frac k{k-0.35}}\left[k-0.35 + 0.35\ln\left(\frac{0.35}{k}\right)\right]}{(k-0.35)^2}$$