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

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

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

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

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derivative of

y = \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}}

\frac{d}{d x} (\frac{1}{4} x^{3} - 2 x)

\frac{d}{d x} (sin (e^{- x}))

x \to \infty lim (\frac{1}{x} + 6)

y^{^{'}} = - 3 (1 - y) (1 - 2 y)

\int_{0}^{24} \frac{x}{9 + 3 x} d x

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