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

We can use quotient rule:

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To find the derivative of ( \frac{2x}{3 + e^x} ), you use the quotient rule.

Let ( u = 2x ) and ( v = 3 + e^x ).

Then, apply the quotient rule formula:

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

Differentiate ( u = 2x ) and ( v = 3 + e^x ):

[ \frac{du}{dx} = 2 ] [ \frac{dv}{dx} = e^x ]

Now, substitute these into the quotient rule formula:

[ \frac{d}{dx} \left( \frac{2x}{3 + e^x} \right) = \frac{(3 + e^x)(2) - (2x)(e^x)}{(3 + e^x)^2} ]

Simplify the expression to get the final derivative.

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To find the derivative of ( \frac{2x}{3+e^x} ), we can use the quotient rule.

The quotient rule states that if you have a function ( \frac{u(x)}{v(x)} ), then the derivative is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

In this case, ( u(x) = 2x ) and ( v(x) = 3+e^x ).

Thus, ( u'(x) = 2 ) and ( v'(x) = e^x ).

Plugging these into the formula, we get:

[ \frac{d}{dx} \left( \frac{2x}{3+e^x} \right) = \frac{2(3+e^x) - 2x(e^x)}{(3+e^x)^2} ]

Simplifying the numerator gives:

[ \frac{6 + 2e^x - 2xe^x}{(3+e^x)^2} ]

Therefore, the derivative of ( \frac{2x}{3+e^x} ) is ( \frac{6 + 2e^x - 2xe^x}{(3+e^x)^2} ).

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