# 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?

-

I would go about finding the inverse of the function and showing that the tangent line at zero has a slope of 0.

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

-
That seems like a good approach. Thanks for your answer! –  Joe S Jun 14 '13 at 0:36

You could find dy/dx, then use that to find when dy/dx is infinity OR doesn't exist, in other words, when the bottom line of the fraction = 0.

I.e, 5x^(2/5)=0, x=0, Therefore, dy/dx doesn't exist when x=0, therefore, point at x=0 has a vertical tangent. When x=0, y=0^(3/5)=0, So, the point with the vertical tangent is (0,0) which is the origin

-
Welcome to Math@SE! Try using LaTeX formatting next time if possible :) –  nbubis Jun 22 at 3:20