Local Linear Approximation; Differentials Solutions To Selected ...

1. Confirm that the stated formula is the local linear approximation at x0 = 0.
(a) (1 + x) 15 ≈ 1 + 15x
Let f(x) = (1 + x) 15 , then f ′ 14 d
(x) = 15(1 + x) dx (x) = 15(1 + x)14 .
Now use the local linear approximation formula for f(x):
(b) tan(x) ≈ x
(c) 1
1+x
f(x) ≈ f(x0) + f ′ (x0)(x − x0)
= (1 + x0) 15 + 15(1 + x0) 14 (x − x0)
= (1 + 0) 15 + 15(1 + 0) 14 (x − 0)
= (1) 15 + 15(1) 14 (x)
= 1 + 15x
Let f(x) = tan(x), then f ′ (x) = sec 2 (x).
f(x) ≈ f(x0) + f ′ (x0)(x − x0)
≈ 1 − x
= tan(x0) + sec 2 (x0)(x − x0)
= tan(0) + sec 2 (0)(x − 0)
= 0 + 1 2 (x)
= x
Let f(x) = 1
1+x = (1 + x)−1 , then f ′ (x) = −1(1 + x) −2 (1) = − 1
(1+x) 2
f(x) ≈ f(x0) + f ′ (x0)(x − x0)
= 1 1
− (x − x0)
1 + x0 (1 + x) 2
= 1
1 + 0 −
1
(x − 0)
(1 + 0) 2
= 1 − x
1