derivative of s(t)= 1/(t+1) — step-by-step solution

We save the definition:

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

To ease differentiation, the expression

\frac{1}{t + 1}

is turned into

{(t + 1)}^{-1}

Let's perform the differentiation:

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

We differentiate using:

\frac{d}{d t} (f^{n}) = n \cdot \frac{d}{d t} (f) \cdot f^{n - 1}

(where f =

t + 1

)

ⓘ The derivative of x^n is n*x^(n-1):

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

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

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

ⓘ Multiplying by -1 gives the opposite:

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

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

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

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

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

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

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

1 + 0

- \frac{1 + 0}{{(t + 1)}^{2}}

ⓘ Adding 0 changes nothing:

1 + 0

1

▶Derivative:

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