Question

1. - / 5.55 points

differentiate.
$y = \frac{t}{(t - 9)^2}$
$y = \square$

Explanation:

Step1: Identify numerator and denominator

Let \( f(t) = t \), \( g(t) = (t-9)^2 \)

Step2: Apply quotient rule

Quotient rule: \( y' = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2} \)

Step3: Compute derivatives

\( f'(t) = 1 \), \( g'(t) = 2(t-9) \)

Step4: Substitute into formula

\( y' = \frac{1 \cdot (t-9)^2 - t \cdot 2(t-9)}{(t-9)^4} \)

Step5: Simplify numerator

Factor out \( (t-9) \): \( (t-9)[(t-9) - 2t] = (t-9)(-t-9) = -(t-9)(t+9) \)

Step6: Cancel common term

Cancel \( (t-9) \): \( y' = \frac{-(t+9)}{(t-9)^3} \)

Answer:

\( -\frac{t+9}{(t-9)^3} \)