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Babolkybya

Answered question

2022-03-17

What is the instantaneous rate of change of

f (x) = x - e^{x^{2} - 7}

at x=3?

Kocura5hs

Beginner2022-03-18Added 5 answers

Instantaneous rate of change is the first derivative f'(x)
From the given

f (x) = x - e^{x^{2} - 7}

The first derivative is

f (x) = x - e^{x^{2} - 7}

f^{'} (x) = 1 - e^{x^{2} - 7} \cdot \frac{d}{d x} (x^{2} - 7)

f^{'} (x) = 1 - e^{x^{2} - 7} \cdot (2 x - 0)

f^{'} (x) = 1 - 2 x \cdot e^{x^{2} - 7}

at x=3

f^{'} (3) = 1 - 2 \cdot (3) \cdot e^{3^{2} - 7}

f^{'} (3) = 1 - 6 \cdot e^{2}

Result:

f^{'} (3) = 1 - 6 \cdot e^{2}

f'(3)=-43.33

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