Question

find $\frac{ds}{dt}$ if $s = \frac{t}{9t + 1}$. $\frac{ds}{dt}=square$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = t$, $u^\prime=1$, $v = 9t + 1$, and $v^\prime=9$.

Step2: Substitute values into quotient - rule

$\frac{ds}{dt}=\frac{1\times(9t + 1)-t\times9}{(9t + 1)^{2}}$.

Step3: Simplify the expression

$\frac{ds}{dt}=\frac{9t + 1-9t}{(9t + 1)^{2}}=\frac{1}{(9t + 1)^{2}}$.

Answer:

$\frac{1}{(9t + 1)^{2}}$