Question 128713
Differentiate by the product rule and then by multiplying then differentiating:
{{{g(x) = (4x-3)(2x^2+3x+5)}}}
First, let {{{u = (4x-3)}}} and {{{v = (2x^2+3x+5)}}}
The product rule is:
{{{dy/dx = u*(dv/dx) + v*(du/dx)}}} Applying this to your problem:
{{{dg/dx = (4x-3)(d(2x^2+3x+5)/dx)+(2x^2+3x+5)(d(4x-3)/dx)}}} Applying the differentiation:
{{{dg/dx = (4x-3)(4x+3)+(2x^2+3x+5)(4)}}} Simplify:
{{{dg/dx = 16x^2+12x-12x-9+8x^2+12x+20}}}
{{{dg/dx = 24x^2+12x+11}}}
Now we'll multiply then differentiate:
{{{(4x-3)(2x^2+3x+5) = 8x^3+12x^2+20x-6x^2-9x-15}}} Simplifying, we get:
{{{g(x) = 8x^3+6x^2+11x-15}}} Now we'll differentiate:
{{{dg/dx = 24x^2+12x+11}}}