Answer
(a) $L(x) = f(a)+ f'(x-a)$
(b) $dy = f'(x)~dx$
(c) We can see an example below that shows the geometric meaning of $\Delta x, dx, \Delta y,$ and $dy$
Work Step by Step
(a) We can write an expression for the linearization of $f$ at $a$:
$L(x) = f(a)+ f'(x-a)$
(b) Suppose $y = f(x)$
Then:
$\frac{dy}{dx} = f'(x)$
$dy = f'(x)~dx$
(c) We can see an example below that shows the geometric meaning of $\Delta x, dx, \Delta y,$ and $dy$
