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
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 $\left(\sin(uv),\frac{1}{w} \cos(uv) \right)\neq (0,0).$
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?
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.
Best Answer
They both satisfy the linear first order differential equation:
$\begin{equation*} f'(t) = A f(t) + B \end{equation*}$
You need at least one other condition, say prove that $x(t_0) = y(t_0)$ for some value $t_0$, or perhaps the same derivative at a point. Note that the solution is
$\begin{equation*} f(t) = c e^{A t} - B/A \end{equation*}$
here $c$ is an unknown constant, to be determined by other conditions.