Answer:
[tex]h'(1) = 23[/tex]
Step-by-step explanation:
Let be [tex]h(x) = (x^{2}+8)\cdot g(x)[/tex], where [tex]r(x) = x^{2} + 8[/tex]. If both [tex]r(x)[/tex] and [tex]g(x)[/tex] are differentiable, then both are also continuous for all x. The derivative for the product of functions is obtained:
[tex]h'(x) = r'(x) \cdot g(x) + r(x) \cdot g'(x)[/tex]
[tex]r'(x) = 2\cdot x[/tex]
[tex]h'(x) = 2\cdot x \cdot g(x) + (x^{2}+8)\cdot g'(x)[/tex]
Given that [tex]x = 1[/tex], [tex]g (1) = -2[/tex] and [tex]g'(1) = 3[/tex], the derivative of [tex]h(x)[/tex] evaluated in [tex]x = 1[/tex] is:
[tex]h'(1) = 2\cdot (1) \cdot (-2) + (1^{2}+8)\cdot (3)[/tex]
[tex]h'(1) = 23[/tex]
