just a quick calculus question here. I'm a bit new to indefinite integrals and I was presented with this problem. Find $f(x)$ if $f''(x) = x^{-2}$. Naturally it is easy to realize that the first derivative $f'(x)$ is simply $-x^{-1}+C_1$, however, when calculating the anti-derivative of $-x^{-1}$ to find $f$, what is the proper way to denote $f(x)$? Because the natural answer is $-x^0$ due to the power rule but that obviously doesn't make any sense.
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If $f(x)=\ln(x)$, then $f'(x)=\frac1x=x^{-1}$.
It seems at first sight surprising, but it easy to see if you are familiar with the formula of the derivate of inverse functions: $$ (g^{-1})'(x)=\frac1{g'(g^{-1}(x))} $$ For $g(x)=e^x$ and $g'(x)=e^x$ we get $g^{-1}(x)=\ln(x)$ hence $$ \ln'(x)=\frac1{e^{\ln(x)}}=\frac1x. $$
Mundron Schmidt
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The antiderivative of $\frac 1x$ is the function whose inverse is exactly equal to its own derivative. Indeed, let $y(x)$ be the antiderivative of $\frac 1x$. Then we have $$\frac{dy}{dx} = \frac 1x$$ Now invert, thinking of the Leibniz notation $\frac{dy}{dx}$ as a rate of change: $$\frac{dx}{dy} = x$$ This means that that $\frac{d}{dx}[x] = x$, i.e. the function x(y) is equal to its own derivative.
This means that the antiderivative $x(y)$ cannot be a polynomial (or rational) function, since such functions are changed when you differentiate them. So $y(x) = \frac 1x$ is then a special case—a non-polynomial—and so the Power Rule doesn't apply here. You can, however, prove that $\frac{d}{dx}[\ln|x|] = \frac 1x$ using the limit definition of the derivative.
