Question

If `G(x)=f(ax+b)` then prove that `G^((n))(x)=a^nf^((n))(ax+b)`?

Answer 1

We want to prove that given:

`G(x)=f(ax+b)` then `G^((n))(x)=a^nf^((n))(ax+b)`

We can prove this assertion by Mathematical Induction

When `n=1` the given result gives:

`G^((1))(x)=a^1f^((1))(ax+b) = af'(ax+b)`

And if we differentiate `G(x)=f(ax+b)` wrt `x` using the chain rule we get:

`G'(x) = f'(ax+b) d/dx(ax+b)`
`" " = f'(ax+b) a`
`" " = af'(ax+b)`

So the given result is true when `n=1`

Now, Let us assume that the given result is true when `n=k`, for some `k in NN`, in which case for this particular value of `k` we have:

`G^((k))(x)=a^kf^((k))(ax+b)`

If we differentiate this wrt `x` using the chain rule we get:

`G^((k+1))(x) = a^kf^((k+1))(ax+b) d/dx(ax+b)`
`" " = a^kf^((k+1))(ax+b) a`
`" " = a^(k+1)f^((k+1))(ax+b)`

Which is the given result with `n=k+1`

So, we have shown that if the given result is true for `n=k` then it is also true for `n=k+1`. But we initially showed that the result was true for `n=1` and so it must also be true for `n=2, n=3, n=4, ...` and so on.

Hence, by the process of mathematical induction the given result is true for all `n in NN^+` QED