Suppose 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 the value of h'(1).
h(x)= [x f(x)] / [ x + g(x) ]
I figured it out myself. Thanks anyways.
To find the value of h'(1), we can differentiate the function h(x) with respect to x and then substitute x=1 into the resulting expression.
Using the quotient rule, h(x) can be written as:
h(x) = [x * f(x)] / [x + g(x)]
Let's differentiate h(x):
h'(x) = [(x + g(x)) * (f'(x) * x + f(x)) - (x * f(x)) * (1 + g'(x))] / [(x + g(x))^2]
Next, substitute x=1 into h'(x):
h'(1) = [(1 + g(1)) * (f'(1) * 1 + f(1)) - (1 * f(1)) * (1 + g'(1))] / [(1 + g(1))^2]
Given that f(1) = 2, f'(1) = -1, g(1) = -2, and g'(1) = 3, we can plug in these values:
h'(1) = [(1 + (-2)) * ((-1) * 1 + 2) - (1 * 2) * (1 + 3)] / [(1 + (-2))^2]
Simplifying further:
h'(1) = [(-1) * 1 - 2 * 4] / [(-1)^2] = [-1 - 8] / 1 = -9
Therefore, the value of h'(1) is -9.