Suppose that f and g are functions such that g(-2) = 2, f(2) = 14, g'(-2) = 1, and f '(2) = -116.
If h(x) = f(g(x)) then the equation of the line tangent to y = h(x) at x=-2 is ?
I don't know how to start this at all. :/
The tangent line is the derivative of h. h = f(g(x)). For x = -2, we just start plugging what we need in.
The chain rule is: f'(g(x))g'(x).
g(x) = g(-2) = 2.
f'(2) = -116
(so that's the first part)
g'(-2) = 1
(and that's the second part)
-116 * 1 = -116
(that's the answer)
Chain rule again. Ignore the h() function, that is not really part of the problem, and can be confusing. First, identify the "links" of the chain you have to deal with.
The function is .
The outer "link" of the chain is and this can be rewritten as simply .
The inner "link" of the chain is the function f(x).
Same as before, we use the chain rule to solve. The derivative of is . Here x = f(x). The problem says that f(-3) = 1, therefore this first part is really just -6.
The second part is the derivative of the inner link, or f'(x). The problem states that f'(-3) = 6. Therefore, the solution is .