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Evaluate the derivative of the function below at the point (6, 0.035). Round to four decimal places. f(x) = 5/(x**2-4 x)**2
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I put it in the exponent form, \[f(x)=5(x^2-4x)^{-2}\]Then, use chain rule, \[f'(x)=5\cdot2\cdot(x^2-4x)^{-3}\cdot(2x-4)=\frac{20x-40}{(x^2-4x)^3}\] Then substitue the x=6, \[f'(6)=0.0463\]
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