3 years ago

Compute the values of dy and Δy for the function y=e^(2x)+6x given x=0 and Δx=dx=0.03.

Mathematics

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

dy = 8·dx
Δy = 0.24

Step-by-step explanation:

The derivative of your function is ...

y' = dy/dx = 2e^(2x) +6

At x=0, the value is ...

y'(0) = 2e^0 +6 = 8

dy = 8·dx

Δy = y'(0)·Δx

Δy = 8(.03)

Δy = 0.24

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