Fourth derivative of $f(x)=e^{4/x}$
Fourth derivative of $f(x)=e^{4/x}$
I keep trying to calculate this but I can't get it right... Can somebody walk me through this? I get it's just the product rule and chain rule, but dang!! The correct answer is supposed to be:
$$f^{(4)}(x)=\frac{4e^{4/x}(24x^3+144x^2+192x+64)}{x^8}$$
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$\begingroup$If you log the function you get the first derivative equal to$$ h_1(x) = f'(x)=-\frac{4}{x^2}e^{\frac{4}{x}} $$The second derivative will be$$ h_2(x) = \frac{8}{x^3}e^{\frac{4}{x}} -\frac{4}{x^2}f'(x) $$And you can use the first derivative. Repeat twice more.
$\endgroup$ $\begingroup$The key rule you need is$$ \frac{d}{dx}\left (\frac{e^{4/x}}{x^n}\right)=\displaystyle - \frac{n e^{4/x}}{x^{n+1}} - \frac{4 e^{4/x}}{x^{n+2}}.$$Then very slowly,\begin{align} &\mathbin{\phantom{=} }\frac{d}{dx}\left (\frac{d}{dx}\left (\frac{d}{dx}\left (\frac{d}{dx}\left (\displaystyle e^{4/x}\right)\right)\right)\right) \\ &=\frac{d}{dx}\left (\frac{d}{dx}\left (\frac{d}{dx}\left (\displaystyle - \frac{4 e^{4/x}}{x^{2}}\right)\right)\right) \\ &= \frac{d}{dx}\left (\frac{d}{dx}\left (\displaystyle \frac{8 e^{4/x}}{x^{3}} + \frac{16 e^{4/x}}{x^{4}}\right)\right) \\ &= \frac{d}{dx}\left (\displaystyle - \frac{24 e^{4/x}}{x^{4}} - \frac{96 e^{4/x}}{x^{5}} - \frac{64 e^{4/x}}{x^{6}}\right) \\ &= \displaystyle \frac{96 e^{4/x}}{x^{5}} + \frac{576 e^{4/x}}{x^{6}} + \frac{768 e^{4/x}}{x^{7}} + \frac{256 e^{4/x}}{x^{8}}. \end{align}
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