Using the following values, find the approximate value of h'(x) at x=1.35

Question

please show your work

h(x)=f(g(x)) is a differentiable function

if x=2.5, f'(x)=3.85

if x=2.7, f'(x)=3.97

if x=2.9, f'(x)=4.03

if x=1.3, g(x)=2.5

if x=1.35, g(x)=2.7

if x=1.40, g(x)=2.9

Answer 1

Question: Using the following values, find the approximate value of h'(x) at x=1.35.

Where h(x) = f(g)(x).

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