Question

Use the figures below to evaluate the indicated derivative, or state that it does not exist. If the derivative does not exist, enter dne in the answer blank. The graph to the left (in black) gives f(x), while the graph to the right gives g(x) (which is constant for values of x

greater than 20).

f(x)

g(x)

ddxf(g(x))|x=15=

Answer 1

We know that d/dx f(g(x)) = f ’(g(x))*g’(x).

So, when x = 15, d/dx f(g(15)) = f ’(g(15))*g’(15)

Now g(15) can be evaluated as 27.5 because the graph of g(x) is part of the line y = .5x + 20 for x between 0 and 20. So let us substitute that in and see what we have got:

d/dx f(g(15)) =f ’(27.5)*g'(15)

Now in order to determine the values of the derivatives at these points, we need to recall that the derivative at a point is equal to the slope at that point. These slopes aren't changing in the immediate vicinity of x = 15 so it is easy enough to collect the value for f '(27) as equal to 1. Meanwhile on the graph of g(x), the slope at x = 15 is .5 so g'(15) = .5

Substituting again we get:

d/dx f(g(15)) = 1*.5 = .5

So the answer is .5