Let $a,b : I \to \mathbb{R}$ be differentiable functions, and $m: I \to \mathbb{R}^2$ be a function such that $m(t)=(a(t),b(t))$.
Further, let there be a differentiable function $f:\mathbb{R}^2\to\mathbb{R}$, and let $z:I\to\mathbb{R}$, $z(t)=f(a(t),b(t))$.
We wish to find $\frac{dz}{dt}$.
We know that $\frac{dz}{dt}=D(z)(t)$. (Total derivative)
So $$D(z)(t)=D(f\circ m)(t)=D(f)(m(t)).D(m)(t)$$ (Chain rule)
$$D(f)(m(t))=[\frac{\partial f}{\partial x}(m(t)) \space\space\space\frac{\partial f}{\partial y}(m(t))]$$
By this I mean, partial derivative of $f$ w.r.t $x$ at $m(t)$.
$$
D(m)(t)=
\left[ {\begin{array}{c}
\frac{da}{dt} \\
\frac{db}{dt} \\
\end{array} } \right]
$$.
Multiplying both gives, $$\frac{dz}{dt}=\frac{\partial f}{\partial x}.\frac{da}{dt}+\frac{\partial f}{\partial y}.\frac{db}{dt}$$
Now how to express this in terms of $z$ only not $f$?
According to the book I have, the answer is $$\frac{dz}{dt}=\frac{\partial z}{\partial a}.\frac{da}{dt}+\frac{\partial z}{\partial b}.\frac{db}{dt}$$
Also, what is the meaning of $\frac{\partial z}{\partial a}$ according to the definition?
Best Answer
The formula
$$\frac{dz}{dt}=\frac{\partial z}{\partial a} \cdot \frac{da}{dt}+\frac{\partial z}{\partial b} \cdot \frac{db}{dt}$$
given in your book does not make sense. In fact, $z$ is a function of one variable $t$ and there are no partial derivatives $\frac{\partial z}{\partial a} ,\frac{\partial z}{\partial b}$ with respect to two variables $a, b$. The correct formula is nothing else than
$$\frac{dz}{dt}=\frac{\partial f}{\partial x} \cdot \frac{da}{dt}+\frac{\partial f}{\partial y} \cdot \frac{db}{dt}$$
or, written pointwise,
$$\frac{dz}{dt}(t_0) =\frac{\partial f}{\partial x}(a(t_0),b(t_0)) \cdot \frac{da}{dt}(t_0)+\frac{\partial f}{\partial y}(a(t_0),b(t_0)) \cdot \frac{db}{dt}(t_0) .$$
It is impossible to eleminate $f$ from this formula. The only "explanation" is that your book abuses notation and writes
$$z(t) = z(a(t),b(t))$$
which is not correct.