Find y' with respect to t for y = 1/(t²+t+1)
Find $y'$ with respect to $t$ for $y = \dfrac{1}{t^{2}+t+1}$.
Explanation of Method
Differentiate the reciprocal of a quadratic.
Step 1: Identify the parts and apply the quotient rule
For $y = \dfrac{u}{v}$ recall $y' = \dfrac{u'v - uv'}{v^{2}}$. Here
\[ u = 1,\quad u' = 0, \qquad v = t^{2}+t+1,\quad v' = 2t+1. \]
Step 2: Substitute and simplify
The first term of the numerator vanishes because $u' = 0$: \[y' = \frac{0\cdot\bigl(t^{2}+t+1\bigr) - 1\cdot(2t+1)}{\bigl(t^{2}+t+1\bigr)^{2}} = -\frac{2t+1}{\bigl(t^{2}+t+1\bigr)^{2}}.\]
Step 3: Verification
\[\text{Reciprocal rule } \left(\frac{1}{v}\right)' = -\frac{v'}{v^{2}} \text{ gives the same expression directly} \;\;\checkmark\]
\[t = 0:\quad y'(0) = -\frac{1}{1} = -1; \qquad \frac{y(0.001)-y(-0.001)}{0.002} \approx -1.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = -\dfrac{2t+1}{\bigl(t^{2}+t+1\bigr)^{2}}\,}\]
Remark
Remark. The quadratic $t^{2}+t+1$ has discriminant $1-4 = -3 \lt 0$, so it is positive for every real $t$ and the function is defined everywhere. The derivative changes sign at $t = -\tfrac12$, where the denominator attains its minimum $\tfrac34$ and $y$ its maximum $\tfrac43$.