# Question #83ab4

##### 2 Answers
Apr 26, 2017

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 $y \left(z\right) = {e}^{2 z}$, we can substitute $y \left(u\right) = {e}^{u}$ where $u = 2 z$. Then,

$\frac{\mathrm{dy}}{\mathrm{dz}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dz}} = {e}^{u} \cdot 2 = 2 {e}^{2 z}$

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

$\frac{{d}^{2} y}{{\mathrm{dz}}^{2}} = 2 \cdot \left(\frac{\mathrm{dy}}{\mathrm{dz}}\right) = 2 \cdot 2 \cdot {e}^{2 z}$

(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

$\frac{{d}^{n} y}{{\mathrm{dz}}^{n}} = {2}^{n} \cdot {e}^{2 z}$

Apr 26, 2017

Use an inductive proof.

Assume that $\frac{{d}^{k} y}{d {z}^{k}} = {2}^{k} {e}^{2 z}$

Then differentiate again:

$\frac{d}{d z} \left(\frac{{d}^{k} y}{d {z}^{k}} = {2}^{k} {e}^{2 z}\right)$

$\implies \frac{{d}^{k + 1} k y}{d {z}^{k + 1}} = {2}^{k + 1} {e}^{2 z}$

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

Next as we are in ${\mathbb{Z}}^{+}$, we set $k = 1$

$\frac{{d}^{1} y}{d {z}^{1}} = {2}^{1} {e}^{2 z}$ is true. So it is true for $k = 1$ and must therefore be true for $k = 2 , 3 , \ldots$ ie $k \in {\mathbb{Z}}^{+}$