Hi, I'm having a lot of trouble understanding this problem. I don't get part iii) at all. Because h (if I'm not wrong) is h: R^2—> R and g : R^3—> R^2
I don't get how am I going to find the directional derivative in the direction of a vector that lives in R^3. Wouldn't I need the vector to be in R^2 ?? And the point, why is it given with three coordinates?
Sorry, but I'm stuck in this problem and I would really appreciate If you could help me solve it.
I would also like to know if : when I have a composition of functions that is not differentiable everywhere, then to find the directional derivative(at a point where the composition is indeed differentiable) , the only way I can do it is with the limit definition, or could I use the dot product between the gradient and the directional vector?
And to do it through the first way, do I need to previously substitute the inner function output as the input of the outer function and then take the limit, or can I do some kind of chain rule or something? For example,
g(t) = (t,1/t)
f(x,y) = x^2 + y
I want h(x,y) = f(g(t))
H is not differentiable, so how can I calculate the directional derivative of h at the point (1,2) in the direction (1,1)?
Thanks a lot !
Best Answer
The function $h = f\circ g $ is a map from $\mathbb R \times \mathbb R \times \mathbb R_0$ to $\mathbb R$ (since $w \neq 0$). First apply $g$ to go from $\mathbb R \times \mathbb R \times \mathbb R_0$ to $\mathbb R^2$ and then apply $f$ to go from $\mathbb R^2$ to $\mathbb R$. You may compute it's expression as follows
$$ f\circ g(u,v,w) = f(\sin(uv),\frac 1 w \cos (uv)) = \frac{(\sin (uv) - \frac 1 w \cos uv)^2}{\sin^2 uv + \frac{1}{w^2}\cos^2(uv)}$$
Notice that for any $(u,v,w)$ we have $\left(\sin(uv),\frac{1}{w} \cos(uv) \right)\neq (0,0).$
If a function $h$ is differentiable at a point $a$ then it is true that the directional derivative in the direction of $v$ is given by the dot product of $v$ and the gradient of $h$ at $a$. If $h$ is not differentiable at $a$ then you should use the limit definition instead.
For your last example with $g(t) = (t,1/t)$ and $f(x,y) = x^2+y$ you wrote
$$ h(x,y) = f \circ g(t)$$
The above makes no sense at all !!!!! What you wrote implies that $f \circ g (t)$ is function of $x,y$ and not of $t$. You should have
$$ f \circ g (t) = f(t,1/t) = t^2 + 1/t$$
Therefore $h = f \circ g$ is a function from $\mathbb R_0 \to \mathbb R$ so speaking of directional derivatives isn't really appropriate here.
I suggest you first try to understand how to compose functions before trying harder problems like these.