find the derivative of a^x from the first principles
Let y = a^x
Then ln(y) = x.ln(a)
Write this as Kx = ln(y), K = ln(a)
Let Δy be the increment in y corresponding to an incremental rise of Δx in x.
Then,
K(x+Δx) = ln(y + Δy)
Kx + KΔx = ln(y + Δy)
KΔx = ln(y + Δy) – Kx = ln(y + Δy) – ln(y) = ln({y + Δy}/y)
KΔx = ln(1 + Δy/y)
Using the expansion of ln(1 + z) = z – z^2/2 + z^3/3 – z^4/4 + …
KΔx =(Δy/y) – (Δy/y)^2/2 + (Δy/y)^3/3 – (Δy/y)^4/4 + …
Dividing both sides by Δy,
KΔx/Δy =(1/y) – (Δy)/(2y^2) + (Δy)^2/(3y^3) – (Δy)^3/(4y^4) + …
In the limit, as the series extends to infinity, and as both Δx and Δy tend to zero, the above expression tends to ...
K.(dx/dy) = 1/y
Rearranging,
dy/dx = K.y
dy/dx = (a^x).ln(a)