Question: Using the following values, find the approximate value of h'(x) at x=1.35.
Where h(x) = f(g)(x).
I'm assuming that you made a typo in your entry of the last line of data. You entered: if x=1.35, g(x)=1.40.
It should be: if x=1.40, g(x)=2.9.
Here we use the chain rule.
If h(x) = f(g) and g() is a function of x, then
dh/dx = (df/dg).(dg/dx)
To find h'(x) at x = 1.35,
we need h'(x) = (df(g)/dg).(dg(x)/dx)
at x = 1.35, g'(x=1.35) = ??
at x = 1.35, g(x=1.35) = 2.7
at g = 2.7, f'(g=2.7) = 3.97
We don't have information on the value(s) of g'(x) for various values of x, but we do have information on the variation of g(x) for different values of x, which shows that g(x) is a straight line function which works out as
g(x) = 4x - 2.7
so, g'(x) = 4. i.e. g'(x=1.35) = 4.
Now we have g'(x=1.35) = 4 and f'(g=2.7) = 3.97,
So h'(x=1.35) = (df(g=2.7)/dg).(dg(x=1.35)/dx) = 3.97*4 = 15.88
Answer: 15.88