What is the derivative of y=(e^{6x})/(e^{6x)+3} ?

The derivative of y=(e^{6x})/(e^{6x)+3} is (18e^{6x})/((e^{6x)+3)^2}

What is the first derivative of y=(e^{6x})/(e^{6x)+3} ?

The first derivative of y=(e^{6x})/(e^{6x)+3} is (18e^{6x})/((e^{6x)+3)^2}

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derivative of

y = \frac{e ^{6 x}}{e ^{6 x} + 3}

\frac{18 e ^{6 x}}{( e ^{6 x} + 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{e ^{6 x}}{e ^{6 x} + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( e ^{6 x} ) ( e ^{6 x} + 3 ) - \frac{d}{d x} ( e ^{6 x} + 3 ) e ^{6 x}}{( e ^{6 x} + 3 ) ^{2}}

\frac{d}{d x} (e^{6 x}) = e^{6 x} \cdot 6

\frac{d}{d x} (e^{6 x} + 3) = 6 e^{6 x}

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

Simplify

\frac{e ^{6 x} \cdot 6 ( e ^{6 x} + 3 ) - 6 e ^{6 x} e ^{6 x}}{( e ^{6 x} + 3 ) ^{2}} : \frac{18 e ^{6 x}}{( e ^{6 x} + 3 ) ^{2}}

= \frac{18 e ^{6 x}}{( e ^{6 x} + 3 ) ^{2}}

\frac{\partial}{\partial x} (e^{- 3 y} cos (π x))

6 y^{^{''}} - y^{^{'}} - 2 y = 0

\int \frac{1 + sin ( x )}{1 + cos ( x )} d x

\frac{\partial}{\partial x} (\frac{( y - 3 x )}{z})

f (x) = x^{2} + 8, (- 3, 17)

What is the derivative of y=(e^{6x})/(e^{6x)+3} ?
The derivative of y=(e^{6x})/(e^{6x)+3} is (18e^{6x})/((e^{6x)+3)^2}
What is the first derivative of y=(e^{6x})/(e^{6x)+3} ?
The first derivative of y=(e^{6x})/(e^{6x)+3} is (18e^{6x})/((e^{6x)+3)^2}