Answer:
18
Step-by-step explanation:
Using the product rule
Given
y = f(x). g(x) , then
[tex]\frac{dy}{dx}[/tex] = f(x). g'(x) + g(x). f'(x)
Here
f(x) = [tex](2x+3)^{5}[/tex] ⇒ f'(x) = 5[tex](2x+3)^{4}[/tex] × [tex]\frac{d}{dx}[/tex] (2x + 3) , then
f'(x) = 10[tex](2x+3)^{4}[/tex]
g(x) = [tex](x+2)^{8}[/tex] ⇒ g'(x) = 8[tex](x+2)^{7}[/tex] × [tex]\frac{d}{dx}[/tex] (x + 2) , then
g'(x) = 8[tex](x+2)^{7}[/tex]
Thus
[tex]\frac{dy}{dx}[/tex] = [tex](2x+3)^{5}[/tex] . 8[tex](x+2)^{7}[/tex] + [tex](x+2)^{8}[/tex] . 10[tex](2x+3)^{4}[/tex] ← factor expression
= [tex](2x+3)^{4}[/tex] [tex](x+2)^{7}[/tex] [ 8(2x + 3) + 10(x + 2)
= [tex](2x+3)^{4}[/tex] [tex](x+2)^{7}[/tex] (16x + 24 + 10x + 20)
= [tex](2x+3)^{4}[/tex] [tex](x+2)^{7}[/tex] (26x + 44)
Thus
[tex]\frac{dy}{dx}[/tex] ( x = - 1)
= [tex]1^{4}[/tex] × [tex]1^{7}[/tex] × 18
= 1 × 1 × 18
= 18
