What is the derivative of ((6e^x)/(2e^x+1)) ?

The derivative of ((6e^x)/(2e^x+1)) is (6e^x)/((2e^x+1)^2)

What is the first derivative of ((6e^x)/(2e^x+1)) ?

The first derivative of ((6e^x)/(2e^x+1)) is (6e^x)/((2e^x+1)^2)

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Popular >

\frac{d}{d x} (\frac{( 6 e ^{x} )}{2 e ^{x} + 1})

\frac{6 e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{( 6 e ^{x} )}{2 e ^{x} + 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 6 \frac{d}{d x} (\frac{e ^{x}}{2 e ^{x} + 1})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 6 \cdot \frac{\frac{d}{d x} ( e ^{x} ) ( 2 e ^{x} + 1 ) - \frac{d}{d x} ( 2 e ^{x} + 1 ) e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

\frac{d}{d x} (e^{x}) = e^{x}

\frac{d}{d x} (2 e^{x} + 1) = 2 e^{x}

= 6 \cdot \frac{e ^{x} ( 2 e ^{x} + 1 ) - 2 e ^{x} e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

Simplify

6 \cdot \frac{e ^{x} ( 2 e ^{x} + 1 ) - 2 e ^{x} e ^{x}}{( 2 e ^{x} + 1 ) ^{2}} : \frac{6 e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

= \frac{6 e ^{x}}{( 2 e ^{x} + 1 ) ^{2}}

\frac{1}{2 x}, at x = 64

\int \frac{1}{x 121 x ^{2} - 121} d x

\int y + x d x

x \to 3 lim (\frac{∣ 3 - x ∣}{x - 3})

(2 x^{3} - 3 x + 1)^{^{'}}

What is the derivative of ((6e^x)/(2e^x+1)) ?
The derivative of ((6e^x)/(2e^x+1)) is (6e^x)/((2e^x+1)^2)
What is the first derivative of ((6e^x)/(2e^x+1)) ?
The first derivative of ((6e^x)/(2e^x+1)) is (6e^x)/((2e^x+1)^2)