What is the derivative of (8e^x)/(9e^x+5) ?

The derivative of (8e^x)/(9e^x+5) is (40e^x)/((9e^x+5)^2)

What is the first derivative of (8e^x)/(9e^x+5) ?

The first derivative of (8e^x)/(9e^x+5) is (40e^x)/((9e^x+5)^2)

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derivative of

\frac{8 e ^{x}}{9 e ^{x} + 5}

\frac{40 e ^{x}}{( 9 e ^{x} + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8 e ^{x}}{9 e ^{x} + 5})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{e ^{x}}{9 e ^{x} + 5})

Apply the Quotient Rule:

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

= 8 \cdot \frac{\frac{d}{d x} ( e ^{x} ) ( 9 e ^{x} + 5 ) - \frac{d}{d x} ( 9 e ^{x} + 5 ) e ^{x}}{( 9 e ^{x} + 5 ) ^{2}}

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

\frac{d}{d x} (9 e^{x} + 5) = 9 e^{x}

= 8 \cdot \frac{e ^{x} ( 9 e ^{x} + 5 ) - 9 e ^{x} e ^{x}}{( 9 e ^{x} + 5 ) ^{2}}

Simplify

8 \cdot \frac{e ^{x} ( 9 e ^{x} + 5 ) - 9 e ^{x} e ^{x}}{( 9 e ^{x} + 5 ) ^{2}} : \frac{40 e ^{x}}{( 9 e ^{x} + 5 ) ^{2}}

= \frac{40 e ^{x}}{( 9 e ^{x} + 5 ) ^{2}}

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What is the derivative of (8e^x)/(9e^x+5) ?
The derivative of (8e^x)/(9e^x+5) is (40e^x)/((9e^x+5)^2)
What is the first derivative of (8e^x)/(9e^x+5) ?
The first derivative of (8e^x)/(9e^x+5) is (40e^x)/((9e^x+5)^2)