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

The derivative of 1/(t+1) is -1/((t+1)^2)

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

The first derivative of 1/(t+1) is -1/((t+1)^2)

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derivative of

\frac{1}{t + 1}

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

Solution steps

\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}}

\int t e^{- 5 t} d t

x \to 5 lim (x^{2} - 8 x + 15)

x \to 0 lim (\frac{2}{x})

f (x) = x^{2} (\frac{2}{x ^{2}} + \frac{5}{x ^{3}})

\int \frac{x ^{2} - 64}{x ^{4}} d x

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