Derivative of e^x and a^x

We know the formula for finding the value of ex
ex = 1 + x + x2/2! + x3/3! + x4/4! + ...

We know the formula for finding the derivative
dy/dx = lim [f(x + δx) - f(x)]/δx
       δx→0


Derivative of ex
y = f(x) = ex = 1 + x + x2/2! + x3/3! + x4/4! + ...
f(x + δx) = ex + δx
Difference
f(x + δx) - f(x) = ex + δx - ex
= exeδx - ex
= ex(eδx - 1)
= ex(1 + δx + δx2/2! + δx3/3! + δx4/4! + ... - 1)
= ex(δx + δx2/2! + δx3/3! + δx4/4! + ...)
= ex(δx(1 + δx/2! + δx2/3! + δx3/4! + δx4/5! + ...))

Ratio
[f(x + δx) - f(x)]/δx
= ex(δx(1 + δx/2! + δx2/3! + δx3/4! + δx4/5! + ...))/δx
= ex(1 + δx/2! + δx2/3! + δx3/4! + δx4/5! + ...)

dy/dx = lim [f(x + δx) - f(x)]/δx
       δx→0

When δx →0 then all the terms containing δx tends to zero and become negligible and
dy/dx = ex


Derivative of ax
y = f(x) = ax = ex loge a
f(x + δx) = e(x + δx)loge a
Difference
f(x + δx) - f(x) = e(x + δx)loge a - ex loge a
= ex loge aeδx loge a - ex loge a
= ex loge a(eδx loge a - 1)
= ex loge a(1 + δx loge a + (δx loge a)2/2! + (δx loge a)3/3! + (δx loge a)4/4! + ... - 1)
= ex loge a(δx loge a + (δx loge a)2/2! + (δx loge a)3/3! + (δx loge a)4/4! + ...)
= ex loge a(δx loge a(1 + (δx loge a)/2! + (δx loge a)2/3! + (δx loge a)3/4! + (δx loge a)4/5! + ...))

Ratio
[f(x + δx) - f(x)]/δx
= ex loge a(δx loge a(1 + (δx loge a)/2! + (δx loge a)2/3! + (δx loge a)3/4! + (δx loge a)4/5! + ...))/δx
= ex loge a(loge a(1 + (δx loge a)/2! + (δx loge a)2/3! + (δx loge a)3/4! + (δx loge a)4/5! + ...))

dy/dx = lim [f(x + δx) - f(x)]/δx
       δx→0

When δx →0 then all the terms containing δx tends to zero and become negligible and
dy/dx = ex loge aloge a
dy/dx = ax loge a

d(ex)/dx = ex
and
d(ax)/dx = ax loge a



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