The Chain Rule. [1 → 2 → 1.]
Theorem: Given $f : R →R^2$ and $g:R^2 →R$.
Suppose $f$ is differentiable at $a$, and $g$ is differentiable at $(x(a), y(a))$ and $P: R →R$ is defined by $P(t) = g(f(t))$. 
Then $P$ is differentiable at $a$ and $P’(a) = g_x ( f (a) ) x’(a) + g _y ( f (a) ) y’(a)$.

 Proof: I. $P(a+h) = g(f(a+h))$
                   $P(a) = g(f(a))$
 II $P(a+h)-P(a) = g(f(a+h))- g(f(a))= g(x(a+h), y(a+h))- g( x(a),y(a))$

III $ \frac{P(a+h)-P(a)} h = \frac {g(f(a+h))- g(f(a))} h = \frac{g(x(a+h), y(a+h))- g( x(a),y(a))} h$
 $^1$ $\approx \frac {g_x (x(a),y(a)) (x(a+h)- x(a)) + g_y (x(a),y(a)) (y(a+h)- y(a))}h$

$= g_x (x(a),y(a)) \frac{(x(a+h)- x(a))} h + g_y (x(a),y(a)) \frac{(y(a+h)- y(a))}h$.  

IV Think:
As $h \to  0, \frac{P(a+h)-P(a)} h \approx g_x (x(a),y(a)) \frac{(x(a+h)- x(a))} h + g_y (x(a),y(a))\frac {(y(a+h)- y(a))}h \to g_x ( f (a) ) x’(a) + g_y ( f (a) ) y’(a)$.
So  $P’(a) = g_x ( f (a) ) x’(a) + g _y ( f (a) ) y’(a)$                  
EOP.
Example: Suppose $f(t) = (\cos(t), 3 \sin(t))$, $g(x,y) = x^2 - y^2$, and $P(t)= g(f(t))$.
Find $P'(t)$.
Solution: $x(t) = \cos(t); y(t) = 3 \sin(t); g_x(x,y) = 2x ; g_y(x,y) = -2y$.
By the Chain Rule  $P’(t) = g_x ( f (t) ) x’(t) + g _y ( f (t) ) y’(t) = 2\cos(t) \cdot (-\sin(t))+ -2 (3\sin(t)) \cdot 3 \cos(t) =  - 20 \sin(t) \cos(t)$ .
 

 $^1$[ $g$ is differentiable at $f(a)$ , $f$ differentiable at $a$]