What is the d/(dt)((Ae^{mt})/(B+Ce^{mt)}) ?

The d/(dt)((Ae^{mt})/(B+Ce^{mt)}) is (me^{mt}AB)/((B+e^{mt)C)^2}

What is the first d/(dt)((Ae^{mt})/(B+Ce^{mt)}) ?

The first d/(dt)((Ae^{mt})/(B+Ce^{mt)}) is (me^{mt}AB)/((B+e^{mt)C)^2}

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

\frac{m e ^{m t} A B}{( B + e ^{m t} C ) ^{2}}

Solution steps

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

Treat

m, A, B, C

as constants

Take the constant out:

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

= A \frac{d}{d t} (\frac{e ^{m t}}{B + C e ^{m t}})

Apply the Quotient Rule:

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

= A \frac{\frac{d}{d t} ( e ^{m t} ) ( B + C e ^{m t} ) - \frac{d}{d t} ( B + C e ^{m t} ) e ^{m t}}{( B + C e ^{m t} ) ^{2}}

\frac{d}{d t} (e^{m t}) = e^{m t} m

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

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

Simplify

A \frac{e ^{m t} m ( B + C e ^{m t} ) - C m e ^{m t} e ^{m t}}{( B + C e ^{m t} ) ^{2}} : \frac{m e ^{m t} A B}{( B + e ^{m t} C ) ^{2}}

= \frac{m e ^{m t} A B}{( B + e ^{m t} C ) ^{2}}

\frac{\partial}{\partial x} (y^{2} (x + y)^{- 2})

\int (tan (\frac{x}{14}))^{5} d x

11 x y^{^{'}} + y = 132 x

(u^{- 2} + u^{- 3}) (u^{5} - 6 u^{2})

\int e^{- x} + 2 d x

What is the d/(dt)((Ae^{mt})/(B+Ce^{mt)}) ?
The d/(dt)((Ae^{mt})/(B+Ce^{mt)}) is (me^{mt}AB)/((B+e^{mt)C)^2}
What is the first d/(dt)((Ae^{mt})/(B+Ce^{mt)}) ?
The first d/(dt)((Ae^{mt})/(B+Ce^{mt)}) is (me^{mt}AB)/((B+e^{mt)C)^2}