Question

differentiate the function.
f(t) = \frac{ - t^2 + 7t + 3 }{ t^2 - 4t + 8 }
f(t) = \square

Explanation:

Step1: Identify numerator and denominator

Let \( u = -t^2 + 7t + 3 \), \( v = t^2 - 4t + 8 \)

Step2: Compute derivatives of u and v

\( u' = -2t + 7 \), \( v' = 2t - 4 \)

Step3: Apply quotient rule

Quotient rule: \( F'(t) = \frac{u'v - uv'}{v^2} \)
Substitute values:
\( u'v = (-2t +7)(t^2 -4t +8) = -2t^3 +8t^2 -16t +7t^2 -28t +56 = -2t^3 +15t^2 -44t +56 \)
\( uv' = (-t^2 +7t +3)(2t -4) = -2t^3 +4t^2 +14t^2 -28t +6t -12 = -2t^3 +18t^2 -22t -12 \)
\( u'v - uv' = (-2t^3 +15t^2 -44t +56) - (-2t^3 +18t^2 -22t -12) = -3t^2 -22t +68 \)
\( v^2 = (t^2 -4t +8)^2 \)

Answer:

\( \frac{-3t^2 -22t +68}{(t^2 -4t +8)^2} \)