Use the definition of a derivative, to find the derivative of the function $g(x)=x^3$ at $x=-1$ .

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Start by recalling the definition of the derivative of a function g(x) at a point x = a, which is given by the limit: $ g'(a) = \lim_{h \to 0} \frac{g(a+h) - g(a)}{h} $.

For the function $ g(x) = x^3 $, we need to find $ g'(x) $ at $ x = -1 $. Substitute $ a = -1 $ into the derivative definition: $ g'(-1) = \lim_{h \to 0} \frac{g(-1+h) - g(-1)}{h} $.

Calculate $ g(-1+h) $ by substituting $ -1+h $ into the function: $ g(-1+h) = (-1+h)^3 $. Expand this expression using the binomial theorem or by direct multiplication: $ (-1+h)^3 = -1 + 3h - 3h^2 + h^3 $.

Calculate $ g(-1) $ by substituting $ -1 $ into the function: $ g(-1) = (-1)^3 = -1 $.

Substitute $ g(-1+h) = -1 + 3h - 3h^2 + h^3 $ and $ g(-1) = -1 $ into the limit expression: $ g'(-1) = \lim_{h \to 0} \frac{(-1 + 3h - 3h^2 + h^3) - (-1)}{h} $. Simplify the expression inside the limit and evaluate the limit as $ h \to 0 $.

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Use the definition of a derivative, to find the derivative of the function g(x)=x3g(x)=x^3g(x)=x3 at x=−1x=-1x=−1.

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Use the definition of a derivative, to find the derivative of the function $g(x)=x^3$ at $x=-1$ .