What is the derivative of (2^x/(2^x-3^x)) ?

The derivative of (2^x/(2^x-3^x)) is (-ln(2)*6^x+ln(3)*6^x)/((2^x-3^x)^2)

What is the first derivative of (2^x/(2^x-3^x)) ?

The first derivative of (2^x/(2^x-3^x)) is (-ln(2)*6^x+ln(3)*6^x)/((2^x-3^x)^2)

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\frac{d}{d x} (\frac{2 ^{x}}{2 ^{x} - 3 ^{x}})

\frac{- ln ( 2 ) \cdot 6 ^{x} + ln ( 3 ) \cdot 6 ^{x}}{( 2 ^{x} - 3 ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 ^{x}}{2 ^{x} - 3 ^{x}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 ^{x} ) ( 2 ^{x} - 3 ^{x} ) - \frac{d}{d x} ( 2 ^{x} - 3 ^{x} ) 2 ^{x}}{( 2 ^{x} - 3 ^{x} ) ^{2}}

\frac{d}{d x} (2^{x}) = 2^{x} ln (2)

\frac{d}{d x} (2^{x} - 3^{x}) = 2^{x} ln (2) - 3^{x} ln (3)

= \frac{2 ^{x} ln ( 2 ) ( 2 ^{x} - 3 ^{x} ) - ( 2 ^{x} ln ( 2 ) - 3 ^{x} ln ( 3 ) ) \cdot 2 ^{x}}{( 2 ^{x} - 3 ^{x} ) ^{2}}

Expand

2^{x} ln (2) (2^{x} - 3^{x}) - (2^{x} ln (2) - 3^{x} ln (3)) \cdot 2^{x} : - ln (2) \cdot 6^{x} + ln (3) \cdot 6^{x}

= \frac{- ln ( 2 ) \cdot 6 ^{x} + ln ( 3 ) \cdot 6 ^{x}}{( 2 ^{x} - 3 ^{x} ) ^{2}}

\frac{d}{d x} (\frac{2 x + 3}{2 x})

\frac{d}{d x} (\frac{x}{sin ( x )})

\frac{\partial ^{2}}{\partial y ^{2}} (sin^{2} (m x + n y))

\int \frac{x ^{2}}{4 x ^{2} + 9} d x

\frac{d}{d x} ((x + 2)^{2})

What is the derivative of (2^x/(2^x-3^x)) ?
The derivative of (2^x/(2^x-3^x)) is (-ln(2)*6^x+ln(3)*6^x)/((2^x-3^x)^2)
What is the first derivative of (2^x/(2^x-3^x)) ?
The first derivative of (2^x/(2^x-3^x)) is (-ln(2)*6^x+ln(3)*6^x)/((2^x-3^x)^2)