If $y=x^4-10$ and if x changes from 2 to 1.99, then the changed value of y is approximately:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5.68
$y=x^4-10$
at $x=2,y=6$
so $Δy=\frac{dy}{dx}×Δx⇒(4x^2)_{x=2}×(0.01)$
$=0.32$
so $y'=y-Δy$
so $6-0.32=5.68$