Using the definition of the derivative, find f^(')(x). f(x)=-x^(2)+6x-7

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Using the definition of the derivative, find  f^(')(x).   f(x)=-x^(2)+6x-7

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Definition of Derivative: The definition of the derivative f'(x) is the limit as h approaches 0 of the difference quotient [f(x+h) - f(x)]/h. Let's apply this definition to the function f(x) = -x^2 + 6x - 7.Calculate f(x+h): First, we need to calculate f(x+h). This means we substitute x+h into the function in place of x: f(x+h) = -(x+h)^2 + 6(x+h) - 7.Expand f(x+h): Now, let's expand f(x+h): f(x+h) = -[(x+h)(x+h)] + 6(x+h) - 7         = -(x^2 + 2xh + h^2) + 6x + 6h - 7.Form Difference Quotient: Next, we form the difference quotient [f(x+h) - f(x)]/h: [f(x+h) - f(x)]/h = [(-x^2 - 2xh - h^2 + 6x + 6h - 7) - (-x^2 + 6x - 7)]/h.Simplify Difference Quotient: We simplify the difference quotient by canceling out the terms that appear in both f(x+h) and f(x): [f(x+h) - f(x)]/h = [-x^2 - 2xh - h^2 + 6x + 6h - 7 + x^2 - 6x + 7]/h                    = [-2xh - h^2 + 6h]/h.Factor Out h: Now, we can factor out h from the numerator: [f(x+h) - f(x)]/h = h(-2x - h + 6)/h.Take Limit: We cancel h from the numerator and denominator: [f(x+h) - f(x)]/h = -2x - h + 6.Finally, we take the limit as h approaches 0: f'(x) = lim(h->0) [-2x - h + 6]        = -2x + 6.

Using the definition of the derivative, find $f'(x)$ . $\newline$ $f(x)=-x^{2}+6x-7$

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Using the definition of the derivative, find f′(x)f'(x)f′(x). \newlinef(x)=−x2+6x−7f(x)=-x^{2}+6x-7f(x)=−x2+6x−7

Using the definition of the derivative, find $f'(x)$ . $\newline$ $f(x)=-x^{2}+6x-7$