What is the d/(dt)(1/(2+t)) ?

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

What is the first d/(dt)(1/(2+t)) ?

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\frac{d}{d z} (\frac{z}{z + \frac{a}{z}})

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\int \frac{sec ( x ) + cos ( x )}{3 cos ( x )} d x

\frac{x}{x ^{2} + 3}