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

`h'(1) = 23`

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`