# How do you differentiate #y=(2e^x-1)/(5e^x+9)#?

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To differentiate the function ( y = \frac{{2e^x - 1}}{{5e^x + 9}} ), you can use the quotient rule:

[ \frac{{d}}{{dx}}\left(\frac{{u}}{{v}}\right) = \frac{{v \cdot \frac{{du}}{{dx}} - u \cdot \frac{{dv}}{{dx}}}}{{v^2}} ]

Where ( u = 2e^x - 1 ) and ( v = 5e^x + 9 ).

The derivatives of ( u ) and ( v ) with respect to ( x ) are:

[ \frac{{du}}{{dx}} = 2e^x \quad \text{and} \quad \frac{{dv}}{{dx}} = 5e^x ]

Now, substitute these into the quotient rule formula:

[ \frac{{d}}{{dx}}\left(\frac{{2e^x - 1}}{{5e^x + 9}}\right) = \frac{{(5e^x + 9) \cdot (2e^x) - (2e^x - 1) \cdot (5e^x)}}{{(5e^x + 9)^2}} ]

Simplify this expression to obtain the derivative of ( y ).

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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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