I have two functions $f:\mathbb{R} \rightarrow \mathbb{R}^2$ , $t \mapsto (t^3,t^2)$ and $g : \mathbb{R}^2 \rightarrow \mathbb{R}$ $(x,y) \mapsto (x^2+y^2)^{\alpha}$

Then we are asked to calculate $ (g \circ f)'(t)$ by using two different methods: First, the chain rule and second: by directly inserting f in g and then just differentiating with respect to t.

After that, we are supposed to say, what the difference is, but unfortunately I get in both cases $(g \circ f)'(t) = \alpha (t^6 + t ^4 )^{\alpha - 1} (6 t^5+ 4 t^3)$

Maybe it is something about the existence of the derivatives...

Unfortunately nobody here left a comment about the choice of $\alpha$. Can anybody comment on this, whether this could be a problem?

  • 1
    $\begingroup$ In general, it is a good thing that different ways to calculate the derivative gets you the same answer. $\endgroup$ – Arthur Oct 28 '13 at 13:18
  • $\begingroup$ I don't see why there should be any difference. Both are legal ways to evaluate the derivative, which you have done correctly. $\endgroup$ – Ross Millikan Oct 28 '13 at 13:20
  • $\begingroup$ My friend, you have been snared by the student's worst nemesis, the trick question $\endgroup$ – TBrendle Oct 28 '13 at 13:24
  • $\begingroup$ I would be glad if you are right, but my point is, that g is not differentiable for all alpha, so maybe, there is something different for some values of alpha... $\endgroup$ – user66906 Oct 28 '13 at 13:26

Daniel Fischer noted in a comment on my now-deleted answer that if $\frac14<\alpha<1$, then $(t^6+t^4)^\alpha$ is differentiable, but $g$ is not, so your concerns were justified.

  • $\begingroup$ :-)...thank you and thanks to Daniel Fischer $\endgroup$ – user66906 Oct 28 '13 at 14:24

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