My question in given an homogeneous of degree 3 function $f(x,y)$ satisfying $f(6,9)=54$, find the value of $g'(2)$ when $g(t)=f(0.5t^2, 0.25t^3+1)$. Can anyone help me with that explaining the concept of the chain rule? Or at least refer me to a good book/site?
Thanks a lot
$$\begin{cases} &\mathbb R\overset{\gamma}{\longmapsto}&\mathbb R^2&\overset{f}{\longmapsto}\mathbb R\\ &t\longmapsto&(\gamma_1(t),\gamma_2(t))&\longmapsto f(\gamma_1(t),\gamma_2(t)) \end{cases}$$
So, $g(t)=f\circ\gamma(t)=f(\gamma_1(t),\gamma_2(t))$
The chain rule can be understood geometrically: The variation for $f$ is its gradient $\nabla f=(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y})$ and it has to be computed along the direction of the curve curve $\gamma$, so is, along its tangent vector $\gamma'(t)=(\gamma_1'(t),\gamma_2'(t))$. So is, $g'(t)$ is the variation of $f$ along the curve $\gamma$.
$g'(t)=\nabla f(\gamma_1(t),\gamma_2(t))\boldsymbol{\cdot}\gamma'(t)=\nabla f(\gamma_1(t),\gamma_2(t))\boldsymbol{\cdot}(\gamma_1'(t),\gamma_2'(t))=(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial x})\boldsymbol{\cdot}(\gamma_1'(t),\gamma_2'(t))$
$g'(t)=\gamma'_1(t)\dfrac{\partial f}{\partial x}+\gamma'_2(t)\dfrac{\partial f}{\partial y}$, with the partials evaluated at $(\gamma_1(t),\gamma_2(t))$
First of all, as the function is homogeneous of degree three $f(3·2,3·3)=3^3f(2,3)=54$, so $f(2,3)=2$
Now $g'(t)=t\dfrac{\partial f}{\partial x}+0.75t^2\dfrac{\partial f}{\partial y}$ with the partials evaluated at $(\gamma_1(t),\gamma_2(t))$
$g'(2)=2\dfrac{\partial f}{\partial x}(2,3)+3\dfrac{\partial f}{\partial y}(2,3)$
But the Euler's Homogeneous Function Theorem states that $x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial x}=nf(x,y)$ being $n$ the degree, $3$ in this case.
$2\dfrac{\partial f}{\partial x}+3\dfrac{\partial f}{\partial x}=3f(2,3)$
Then $g'(2)=2\dfrac{\partial f}{\partial x}+3\dfrac{\partial f}{\partial x}=3f(2,3)=3·2=6$