Use the product rule: f'(x) =(x³+2x+3)'(3x³-6x²-8x+1) + (x³+2x+3)(3x³-6x²-8x+1)' f'(x) =(3x²+2)(3x³-6x²-8x+1) + (x³+2x+3)(6x²-12x-8)
Use the product rule again: f''(x) = (6x)(3x³-6x²-8x+1) + (3x²+2)(9x²-12x-8) + (3x²+2)(6x²-12x-8) + (x³+2x+3)(12x-12)
We only care about the coefficient of the x² term so let's extract the operations of terms that give us x²: (6x)(-8x)+(3x²)(-8)+2(9x²)+(3x²)(-8)+2(6x²)+(2x)(12x) = - 48x²+(-24x²)+18x²+(-24x²)+12x²+24x² = 54x²
The diagonal of any square always occurs in a ratio of relative to the square's sides. That means if is the side length, then gives the length of the diagonal. It follows that .
relative to the square's sides. That means if 
is the side length, then 
gives the length of the diagonal. It follows that 
.
square inches