Question

Let g be a function such that g(9)=0 and g'(9)=2 let h be the function h(x)=square root of x

evaluate d/dx[g(x)*h(x)] at x=9

Answer 1

Answer: 6

Work Shown:

First we'll need the derivative of h(x)

`h(\text{x}) = \sqrt{\text{x}}\\\\h(\text{x}) = \text{x}^(1/2)\\\\h'(\text{x}) = (1/2)\text{x}^(-1/2)\\\\h'(\text{x}) = \frac{1}{2\text{x}^(1/2)}\\\\h'(\text{x}) = \frac{1}{2\sqrt{\text{x}}}\\\\`

Then let f(x) = g(x)*h(x)

Use the product rule to evaluate f ' (9).

`f(\text{x}) = g(\text{x})*h(\text{x})\\\\f'(\text{x}) = \frac{d}{d\text{x}}\left[g(\text{x})*h(\text{x})\right]\\\\f'(\text{x}) = g'(\text{x})*h(\text{x}) + g(\text{x})*h'(\text{x})\\\\f'(\text{x}) = g'(\text{x})*\sqrt{\text{x}} + g(\text{x})*\frac{1}{2\sqrt{\text{x}}}\\\\f'(9) = g'(9)*√(9) + g(9)*(1)/(2√(9))\\\\f'(9) = 2*√(9) + 0*(1)/(2√(9))\\\\f'(9) = 2*3 + 0\\\\f'(9) = 6\\\\`