What is the derivative of (xe^t)/(1+xe^t) ?

The derivative of (xe^t)/(1+xe^t) is (e^t)/((1+e^tx)^2)

What is the first derivative of (xe^t)/(1+xe^t) ?

The first derivative of (xe^t)/(1+xe^t) is (e^t)/((1+e^tx)^2)

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derivative of

\frac{x e ^{t}}{1 + x e ^{t}}

\frac{e ^{t}}{( 1 + e ^{t} x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x e ^{t}}{1 + x e ^{t}})

Treat

t

as a constant

Take the constant out:

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

= e^{t} \frac{d}{d x} (\frac{x}{1 + x e ^{t}})

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

\frac{d}{d x} (1 + x e^{t}) = e^{t}

= e^{t} \frac{1 \cdot ( 1 + x e ^{t} ) - e ^{t} x}{( 1 + x e ^{t} ) ^{2}}

Simplify

e^{t} \frac{1 \cdot ( 1 + x e ^{t} ) - e ^{t} x}{( 1 + x e ^{t} ) ^{2}} : \frac{e ^{t}}{( 1 + e ^{t} x ) ^{2}}

= \frac{e ^{t}}{( 1 + e ^{t} x ) ^{2}}

h \to \infty lim (\frac{2}{h} + 3)

4 x^{4} - 4 x^{2}, 5 x^{2}, 0, 1.5

\frac{d}{d x} (sec (x) - \frac{1}{x ^{3}} + 5)

x \to 0 lim (x ln (tan (2 x)))

What is the derivative of (xe^t)/(1+xe^t) ?
The derivative of (xe^t)/(1+xe^t) is (e^t)/((1+e^tx)^2)
What is the first derivative of (xe^t)/(1+xe^t) ?
The first derivative of (xe^t)/(1+xe^t) is (e^t)/((1+e^tx)^2)