Question

find the derivative of the function. f(x) = (2x - 9)^4(x^2 + x + 1)^5 f(x) =

Explanation:

Step1: Apply product rule

Let $u=(2x-9)^4$, $v=(x^2+x+1)^5$. Product rule: $f'(x)=u'v + uv'$.

Step2: Compute $u'$ using chain rule

$u' = 4(2x-9)^3 \cdot 2 = 8(2x-9)^3$.

Step3: Compute $v'$ using chain rule

$v' = 5(x^2+x+1)^4 \cdot (2x+1) = 5(2x+1)(x^2+x+1)^4$.

Step4: Substitute into product rule

$f'(x)=8(2x-9)^3(x^2+x+1)^5 + (2x-9)^4 \cdot 5(2x+1)(x^2+x+1)^4$.

Step5: Factor common terms

Common terms: $(2x-9)^3(x^2+x+1)^4$.
$f'(x)=(2x-9)^3(x^2+x+1)^4[8(x^2+x+1) + 5(2x+1)(2x-9)]$.

Step6: Simplify the bracket

$8(x^2+x+1) = 8x^2 + 8x + 8$
$5(2x+1)(2x-9)=5(4x^2 -18x +2x -9)=5(4x^2 -16x -9)=20x^2 -80x -45$
Sum: $8x^2+8x+8+20x^2-80x-45=28x^2 -72x -37$.

Answer:

$(2x - 9)^3(x^2 + x + 1)^4(28x^2 - 72x - 37)$