Find y' for y = (x²-2x+3)^15
Find $y'$ for $y = \bigl(x^{2}-2x+3\bigr)^{15}$.
Explanation of Method
A high power of a quadratic — the standard first chain-rule exercise.
Step 1: Apply the chain rule
For $y = u^{n}$ the chain rule gives $y' = nu^{n-1}\cdot u'$. Here
\[ u = x^{2}-2x+3, \qquad n = 15, \qquad u' = 2x-2. \]
Expanding the fifteenth power first would be possible but absurd; the chain rule avoids it entirely.
Step 2: Substitute and factor
\[ y' = 15\bigl(x^{2}-2x+3\bigr)^{14}\cdot(2x-2) = 30(x-1)\bigl(x^{2}-2x+3\bigr)^{14}. \]
Step 3: Verification
\[x = 1:\quad u = 1-2+3 = 2 \;\Longrightarrow\; y' = 30\cdot 0\cdot 2^{14} = 0, \text{ a critical point} \;\;\checkmark\]
\[x = 0:\quad y' = 30(-1)\cdot 3^{14} = -30\cdot 4\,782\,969 = -143\,489\,070 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 30(x-1)\bigl(x^{2}-2x+3\bigr)^{14}\,}\]
Remark
Remark. The inner quadratic has discriminant $4-12 = -8 \lt 0$, so $x^{2}-2x+3 = (x-1)^{2}+2 \geq 2$ is never zero: the function is defined and positive for every real $x$. The sign of $y'$ is therefore the sign of $x-1$, giving a global minimum $y(1) = 2^{15} = 32\,768$.