Question

Question #83ab4

Answer 1

Yes, this is true. See the explanation below...

Explanation:

This answer many not be as formal as a proper mathematical proof, but I hope it will do.

According to the chain rule, when we differentiate `y(z)=e^(2z)`, we can substitute `y(u)=e^u` where `u=2z`. Then,

`(dy)/(dz) = (dy)/(du) * (du)/dz = e^u * 2 = 2 e^(2z)`

When we differentiate a second time. the process repeats itself, as the differentiation of the exponential yields another factor of 2:

`(d^2y)/(dz^2) = 2* ((dy)/(dz)) = 2*2*e^(2z)`

(In truth, this is an application of the product rule, but I have omitted the second term, as the derivative of 2 is zero.)

This pattern continues for as many times as you choose to do the differentiation. Each derivative places another factor of 2 in front of the result, so that after `n` differentiations, we have

`(d^ny)/(dz^n) = 2^n * e^(2z)`

Answer 2

Use an inductive proof.

Assume that `(d^k y )/(d z^ k) = 2^k e^(2z)`

Then differentiate again:

`(d )/(d z) ((d^k y )/(d z^ k) = 2^k e^(2z))`

`implies (d^(k+1)k y )/(d z^( k+1)) = 2^(k+1) e^(2z)`

So if it is true for `k`, it must be true for `k + 1`

Next as we are in `ZZ^+`, we set `k = 1`

`(d^(1) y )/(d z^(1)) = 2^1 e^(2z)` is true. So it is true for `k = 1` and must therefore be true for `k = 2,3,...` ie `k in ZZ^+`