Question #83ab4

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)#

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^+#