Find y' for y = (ax²+bx+c)^n, where a, b, c, n are
Find $y'$ for $y = \bigl(ax^{2}+bx+c\bigr)^{n}$, where $a$, $b$, $c$, $n$ are constants.
Explanation of Method
The general form behind Problem 82.
Step 1: Apply the chain rule
For $y = u^{n}$ the chain rule gives $y' = nu^{n-1}\cdot u'$. Here the inner function is the general quadratic, whose derivative was found in Problem 29: $u = ax^{2}+bx+c, \qquad u' = 2ax+b$.
Step 2: Substitute
\[ y' = n\bigl(ax^{2}+bx+c\bigr)^{n-1}\cdot(2ax+b). \]
Step 3: Verification
\[a=1,\ b=-2,\ c=3,\ n=15:\quad y' = 15(2x-2)\bigl(x^{2}-2x+3\bigr)^{14}, \text{ Problem 82} \;\;\checkmark\]
\[n = 1:\quad y' = 1\cdot(2ax+b)\bigl(ax^{2}+bx+c\bigr)^{0} = 2ax+b, \text{ Problem 29} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = n(2ax+b)\bigl(ax^{2}+bx+c\bigr)^{n-1}\,}\]
Remark
Remark. For $a \neq 0$ the factor $2ax+b$ vanishes at $x = -\dfrac{b}{2a}$, the vertex of the inner parabola — so whatever the exponent $n$, the composite function always has a stationary point directly above that vertex. For non-integer $n$ the formula additionally requires $ax^{2}+bx+c \gt 0$.