then `h` is differentiable at `a` and `h'(a) = f'(g(a)) cdot g'(a)`.

Let `b = g(a)`and `k = g(a+h) - g(a) = g(x) - g(a)` for `h ne 0`.

Note `g(x) = g(a+h) = g(a) + k = b + k`. See Figure 1 .

From the assumption that `g` is differentiable at `a`, we have that `g` is also continuous at `a` . Thus we can conclude that as `h to 0, k to 0`.

We’ll follow the usual steps in finding the derivative of P at a:

Step I: `P(a+h) = f(g(a+h))`

`\underline{- P(a)\ \ \ \ = f(g(a))}`

Step II: `P(a+h) - P(a) = f(g(a+h)) - f(g(a)) = f(b + k) - f(b)`.

Now we assumed that `k ne 0` . [Note: This is a major assumption for some functions.]

So

`P(a+h) - P(a) = {f(b + k) - f(b)}/k . k`

Therefore

`P'(a) = lim_{h to 0} { P(a+h) - P(a)}/h`

`= lim _{h to 0, k to 0} {f(b + k) - f(b)}/k cdot k/h`

`= lim _{h to 0, k to 0} {f(b + k) - f(b)}/k cdot {g(a+h)-g(a) }/h`

`= f '(b) cdot g'(a)`

`= f '(g(a)) cdot g'(a)`.

Suppose that `k = 0` for values of `h` arbitrarily close to `0`.

Since we assume that g is differentiable we know that

`lim_{h to 0} {g(a+h)-g(a)}/ h` must exist. Our assumption that `k = 0` for `h` arbitrarily close to `0` means that

there is a sequence of values of ` h`, {`h_n`} with `h_n to 0` and `g(a+h_n) -g(a) = 0` for all `n`.

Thus `lim_{n to oo} {g(a+h_n)-g(a)}/ {h_n } = lim_{n to oo} 0/ {h_n }= 0` . [See Figure 2 ]

Thus `g'(a) = lim_{ h to 0} {g(a+h)-g(a)} / h = 0`. [0 is the only possible limit.]

To complete the argument we need only show that `P'(a)= 0`.

But for precisely the same h values that had `k = g(a+h)-g(a) = 0`, we have `g(a+h) = g(a)`. Thus for these values of `h`

`P(a+h) - P(a) = f(g(a+h)) - f(g(a)) = f(b + k) - f(b) = f(b) - f(b) = 0.`

and hence `{P(a+h) - P(a)}/h = 0`. [See Figure 3]

Now for any `h` where `k ne 0`, see Figure 1, the argument of part I is still valid to show that `{P(a+h) - P(a)}/h to 0` as `h to 0`.

[This is primarily because `g'(a)=0`.]

In summary then , as h approaches 0 either `{P(a+h)-P(a)}/h` is

close to or actually is 0.

Thus `P'(a) =lim_{h to 0} {P(a+h) - P(a) }/h= 0`. EOP.