# How do you find the derivative of #f(x)=(x^2-5x+2)(x-2/x)#?

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To find the derivative of ( f(x) = (x^2 - 5x + 2)\left(\frac{x - 2}{x}\right) ), you can apply the product rule. The product rule states that if ( u(x) ) and ( v(x) ) are differentiable functions of ( x ), then the derivative of ( u(x)v(x) ) with respect to ( x ) is ( u'(x)v(x) + u(x)v'(x) ).

Let ( u(x) = x^2 - 5x + 2 ) and ( v(x) = \frac{x - 2}{x} ). Then, ( u'(x) = 2x - 5 ) and ( v'(x) = \frac{(x)(1) - (x - 2)(1)}{x^2} = \frac{2}{x^2} ).

Applying the product rule, the derivative of ( f(x) ) is:

[ f'(x) = u'(x)v(x) + u(x)v'(x) ] [ f'(x) = (2x - 5)\left(\frac{x - 2}{x}\right) + (x^2 - 5x + 2)\left(\frac{2}{x^2}\right) ] [ f'(x) = \frac{(2x - 5)(x - 2)}{x} + \frac{(x^2 - 5x + 2)(2)}{x^2} ] [ f'(x) = \frac{2x^2 - 9x + 10}{x} + \frac{2x^2 - 10x + 4}{x^2} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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