What is the derivative of A/(B+Ce^x) ?

The derivative of A/(B+Ce^x) is -(e^xAC)/((B+e^xC)^2)

What is the first derivative of A/(B+Ce^x) ?

The first derivative of A/(B+Ce^x) is -(e^xAC)/((B+e^xC)^2)

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\frac{d}{d x} (\frac{A}{B + C e ^{x}})

- \frac{e ^{x} A C}{( B + e ^{x} C ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{A}{B + C e ^{x}})

Treat

A, B, C

as constants

Take the constant out:

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

= A \frac{d}{d x} (\frac{1}{B + C e ^{x}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= A \frac{d}{d x} ((B + C e^{x})^{- 1})

Apply the chain rule:

- \frac{1}{( B + C e ^{x} ) ^{2}} \frac{d}{d x} (B + C e^{x})

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

\frac{d}{d x} (B + C e^{x}) = C e^{x}

= A (- \frac{1}{( B + C e ^{x} ) ^{2}} C e^{x})

Simplify

A (- \frac{1}{( B + C e ^{x} ) ^{2}} C e^{x}) : - \frac{e ^{x} A C}{( B + e ^{x} C ) ^{2}}

= - \frac{e ^{x} A C}{( B + e ^{x} C ) ^{2}}

\int e^{a x} d x

f (x) = \frac{1 - x ^{2}}{2 x + 1}

2 t x^{^{''}} - x^{^{'}} + \frac{1}{( x ^{^{'}} )} = 0

\int \frac{( x ^{3} )}{x ^{2} + 16} d x

\int \frac{4 + 6 u}{u} d u

What is the derivative of A/(B+Ce^x) ?
The derivative of A/(B+Ce^x) is -(e^xAC)/((B+e^xC)^2)
What is the first derivative of A/(B+Ce^x) ?
The first derivative of A/(B+Ce^x) is -(e^xAC)/((B+e^xC)^2)