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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

t \to 0 lim (\frac{sin ( k t )}{t})

f (x) = \frac{x}{a}

f (x) = x^{2} (4 - x)^{2}, at x = 1

\int_{0}^{1} \frac{1}{1 - x ^{2}} d x