# calculus

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

(2.001)^5

• calculus -

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)

• calculus -

approximate 4 square root of 15.8

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