Why $\Delta y \neq \cos((2.03)^2+1)-2-(\cos(2^2+1)-2)$?

Question

On computing $\Delta y$ from $x=2$ to $x=2.03$:

If $\Delta y = f(x+\Delta x) -f(x)$ and $y=\cos(x^2+1)-x$ why

$\Delta y \neq \cos((2.03)^2+1)-2-(\cos(2^2+1)-2)$ ?

Asumming $\Delta x=0.03$ and $x=2$

Original solution is:

$\Delta y = \cos((2.03)^2+1)-2.03-(\cos(2^2+1)-2)$

This implies:

$\Delta x =0.03$ and $x=2=2.03$?!

Context: http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.aspx (Example 2)

Thanks in advance.

Answer 1

We have $y = f(x) = \cos(x^2 + 1) - x$ and thus

$$\Delta y = f(x+\Delta x) -f(x) = f(2.03) - f(2) = \left(\cos((2.03)^2+1)-\color{red}{2.03}\right)-(\cos(2^2+1)-2)$$