You are trying to prove by induction that $\displaystyle \frac{d^n}{dx^n}\bigl(x^{n-1}e^{1/x}\bigr) = (-1)^n\frac{e^{1/x}}{x^{n+1}}$. So you assume the result for n=k, and you are trying to prove it for n=k+1 by differentiating one more time. Where you are going wrong is that you are finding the (k+1)th derivative of the same function. But the result for n=k+1 refers to the (k+1)th derivative of the function $\displaystyle x^{k}e^{1/x} = x\bigl(x^{k-1}e^{1/x}\bigr)$. You can differentiate that k+1 times by regarding it as a product (as I just indicated) and using Leibniz's rule together with the inductive hypothesis.
$\displaystyle x^{k}e^{1/x} = x\bigl(x^{k-1}e^{1/x}\bigr)$
so i made the k+1 derivative and i need to multiply both sides by the n+1 derivative of x
in order to make it the k+1 expression
??
so i need to do the derivative of x*f(x)=1*f(x)+x*f'(x)
of the left side
i have both f(x) and f'(x)
so if i will substitute
it will work??
it didnt work
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