# Prove that the graph $y=x^{3/5}$ has a vertical tangent at the origin.

My first attempt was to just differentiate and get $y'=\frac{3}{5x^{2/5}}$ and the at the origin, the gradient is $\infty$, but I'm not sure if this is sufficient enough.

Then I differentiated using the definition of the derivative as a limit to try and make it more formal but I'm still not sure if that suffices.

Any ideas on how to 'prove' this properly?

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$$y=f(x)=x^{3/5}$$ $$f^{-1}(x)=x^{5/3}$$ $$\frac{df^{-1}}{dx}=\frac{5}{3}x^{2/3}$$ $${\frac{df^{-1}}{dx}}_{|x=0}=0$$