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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 1 + lim (\frac{6}{x ^{3} - 1})

\int x^{2} - x^{3} d x

\int_{0}^{1} x^{3} e^{x^{2}} d x

\int arctan (x y) d x

y = \frac{5}{x}, y = \frac{10}{x ^{2}}, x = 6

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