Question: "What is the instantaneous rate of change of..."

Ramon Powell

Answered question

2022-03-17

What is the instantaneous rate of change of

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

at x=2?

Heidy David

Beginner2022-03-18Added 5 answers

The instantaneous rate of change is the derivative of our function (which is the rate of change) evaluated at the specific point (at x=2).
So we get:

f^{'} (x) = (2 x) e^{x} + (x^{2} - 2) e^{x}

using the Product Rule.
Evaluate it at x=2 to get:

f^{'} (2) = 4 e^{2} + 2 e^{2} = 6 e^{2} = 44.334

Camryn Heath

Beginner2022-03-19Added 4 answers

Apply the product rule:

\frac{d}{d x} f (x) g (x) = f (x) \frac{d}{d x} g (x) + g (x) \frac{d}{d x} f (x)

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

; to find

\frac{d}{d x} f (x)

:
Differentiate

x^{2} - 2

term by term:
Apply the power rule:

x^{2}

goes to 2x
The derivative of the constant (-1)2 is zero.
The result is: 2x

g (x) = e^{x}

; to find

\frac{d}{d x} g (x) :

The derivative of

e^{x}

it itself.
The result is:

2 x e^{x} + (x^{2} - 2) e^{x}

Now simplify:

(x^{2} + 2 x - 2) e^{x}

The answer is:

(x^{2} + 2 x - 2) e^{x}

By putting the value of x, we get:
44

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