Question

Suppose that f and g are functions that are differentiable at x = 1 and that f(1) = 2, f '(1) = −1, g(1) = −2, and g'(1) = 3. Find h'(1). h(x) = (x2 + 8)g(x)

Answer 1

Answer:

$h'(1) = 23$

Step-by-step explanation:

Let be $h(x) = (x^{2}+8)\cdot g(x)$, where $r(x) = x^{2} + 8$. If both $r(x)$ and $g(x)$ are differentiable, then both are also continuous for all x. The derivative for the product of functions is obtained:

$h'(x) = r'(x) \cdot g(x) + r(x) \cdot g'(x)$

$r'(x) = 2\cdot x$

$h'(x) = 2\cdot x \cdot g(x) + (x^{2}+8)\cdot g'(x)$

Given that $x = 1$, $g (1) = -2$ and $g'(1) = 3$, the derivative of $h(x)$ evaluated in $x = 1$ is:

$h'(1) = 2\cdot (1) \cdot (-2) + (1^{2}+8)\cdot (3)$

$h'(1) = 23$