What is the d/(dt)(t/(1+t)) ?

The d/(dt)(t/(1+t)) is 1/((1+t)^2)

What is the first d/(dt)(t/(1+t)) ?

The first d/(dt)(t/(1+t)) is 1/((1+t)^2)

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\frac{d}{d t} (\frac{t}{1 + t})

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

Solution steps

\frac{d}{d t} (\frac{t}{1 + t})

Apply the Quotient Rule:

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

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

\frac{d t}{d t} = 1

\frac{d}{d t} (1 + t) = 1

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

Simplify

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

n = 1 \sum \infty \frac{1}{n e ^{n}}

\frac{d ^{6}}{d x ^{6}} (x^{- 6})

\frac{\partial}{\partial x} ((x^{2} - a - 1) (x^{2} + a - 1))

\frac{\partial}{\partial x} (3 x^{2} + y^{2})

y = 3 x^{2} + e^{2} x + π^{2}