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How do you find the derivative of #(x-3) /( 2x+1)#?

Answer 1

Christopher Allers

#d/dx ( (x-3)/(2x+1)) = 7/(2x+1)^2#

Using the quotient rule:

#d/dx ( (x-3)/(2x+1)) = ( (2x+1) d/dx (x-3) - (x-3)d/dx (2x+1))/(2x+1)^2#

#d/dx ( (x-3)/(2x+1)) = ( (2x+1) - 2(x-3))/(2x+1)^2#

#d/dx ( (x-3)/(2x+1)) = ( 2x+1- 2x+6)/(2x+1)^2#

#d/dx ( (x-3)/(2x+1)) = 7/(2x+1)^2#

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Answer 2

Luke Alvarez

To find the derivative of ((x-3)/(2x+1)), you can use the quotient rule. The quotient rule states that if you have a function of the form (f(x) = \frac{u(x)}{v(x)}), then the derivative of (f(x)) with respect to (x) is given by:

[f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}]

Where (u(x)) and (v(x)) are differentiable functions, and (u'(x)) and (v'(x)) are their respective derivatives.

So applying the quotient rule to ((x-3)/(2x+1)), you get:

[f'(x) = \frac{(1)(2x+1) - (x-3)(2)}{(2x+1)^2}]

Simplify the expression:

[f'(x) = \frac{2x+1 - 2x + 6}{(2x+1)^2}]

[f'(x) = \frac{7}{(2x+1)^2}]

So, the derivative of ((x-3)/(2x+1)) is (\frac{7}{(2x+1)^2}).

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Answer from HIX Tutor

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.