f(t)=(t^3, |t|^3) is a parametric representation of y=f(x)=|x|.(adsbygoogle = window.adsbygoogle || []).push({});

Consider y=|x|,

the left hand derivative f '-(0)=-1 and the right hand derivative f '+(0)=1, so f(x) is clearly not differentiable at 0.

But

f '(t)=(3t^2, 3t^2) for t>=0

f '(t)=(3t^2, -3t^2) for t<=0

f '(0)=(0,0) and f(t) is differentiable at 0 (my textbook says this explicitly)

These 2 are talking about the same point, how can one gives that it's differentiable at 0 and the other gives that it's not differentiable at 0?

==============================================

Secondly, my textbook says that f '(a)>0 (a is real number) doesn't imply that f is increasing in some neighbourhood of a, how come?

Thanks for explaining!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Differentiability and parametric curves

**Physics Forums | Science Articles, Homework Help, Discussion**