If $y=x^4-10$ and if x changes from 2 to 1.99, than the approximate change in y is

-0.32

Let $x=2, x+\Delta x=1.99$. Then, $\Delta x=1.99-2=-0.01$

Let $d x=\Delta x=-0.01$

We have,

$y=x^4-10 \Rightarrow \frac{d y}{d x}=4 x^3 \Rightarrow\left(\frac{d y}{d x}\right)_{x=2}=4(2)^3=32$

Now, $d y=\frac{d y}{d x} d x$

$\Rightarrow d y=32(-0.01)=-0.32$

$\Rightarrow \Delta y=-0.32$ approximately              [∵ Δy ≅ dy]

So, approximate change in $y=-0.32$