Find the derivative of the function. $$ s(t)=\frac{1}{t^{2}+6 t-5} $$

Question

UpStudy AI · Accepted Answer

To find the derivative of the function $$ s(t) = \frac{1}{t^{2} + 6t - 5} $$, we can use the chain rule for differentiation.

Let $$ f(t) = t^{2} + 6t - 5 $$. Then, $$ s(t) = \frac{1}{f(t)} $$.

The derivative of $$ s(t) $$ with respect to $$ t $$ is:

$$
s'(t) = \frac{d}{dt} \left( \frac{1}{f(t)} \right) = -\frac{f'(t)}{(f(t))^2}
$$

First, compute $$ f'(t) $$:

$$
f'(t) = \frac{d}{dt} (t^{2} + 6t - 5) = 2t + 6
$$

Now, substitute $$ f(t) $$ and $$ f'(t) $$ into the derivative formula:

$$
s'(t) = -\frac{2t + 6}{(t^{2} + 6t - 5)^2}
$$

**Final Answer:**

After simplifying, the derivative is –(2 t + 6) divided by (t² + 6 t – 5) squared. Thus,

s ′(t) = −(2t+6)/(t²+6t−5)²
The derivative of $$ s(t) = \frac{1}{t^{2} + 6t - 5} $$ is $$ s'(t) = -\frac{2t + 6}{(t^{2} + 6t - 5)^2} $$.

Find the derivative of the function. \[ s(t)=\frac{1}{t^{2}+6 t-5} \]