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For the following equation, what is the instantaneous rate of change at
x=-1?

f(x)=x^(5)+4x^(2)+4
Answer:

For the following equation, what is the instantaneous rate of change at $x=-1 ?$ $\newline$ $f(x)=x^{5}+4 x^{2}+4$ $\newline$ Answer:

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Find derivatives using logarithmic differentiation

Full solution

Q. For the following equation, what is the instantaneous rate of change at

x=-1 ?

\newline

f(x)=x^{5}+4 x^{2}+4

\newline

Answer:

Calculate Derivative: To find the instantaneous rate of change of the function $f(x)$ at $x=-1$ , we need to calculate the derivative of $f(x)$ and then evaluate it at $x=-1$ . The function given is $f(x) = x^5 + 4x^2 + 4$ . We will find the derivative $f'(x)$ using the power rule, which states that the derivative of $x^n$ is $n\cdot x^{(n-1)}$ .
Apply Power Rule: Applying the power rule to each term of $f(x)$ , we get: $\newline$ $f'(x) = \frac{d}{dx} (x^5) + \frac{d}{dx} (4x^2) + \frac{d}{dx} (4)$ $\newline$ $f'(x) = 5x^4 + 8x + 0$ $\newline$ The derivative of a constant is $0$ , so the last term disappears after differentiation.
Evaluate at $x=-1$ : Now we will evaluate the derivative at $x=-1$ to find the instantaneous rate of change at that point. $\newline$ $f'(-1) = 5(-1)^4 + 8(-1)$ $\newline$ $f'(-1) = 5(1) - 8$ $\newline$ $f'(-1) = 5 - 8$ $\newline$ $f'(-1) = -3$

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