Use a linear approximation (or differentials) to estimate the given number

Question

(2.001)^5

Answer 1

Let y=x⁵,
y'=5x⁴.
y(x+h)=y(x)+y'(2)*h (approximately)
Therefore, an approximation to y(2.001) is
y(2.001)
=y(2)+y'(2)*0.001
=2^5 + 5(2)^4*0.001 (approximately)

Answer 2

approximate 4 square root of 15.8