Differentiate a^x with respect to x

y=a^x

take natural logs (also written as ln or log base e) of both sides

lny=lna^x

by logarithms rules lna^x=xlna

lny=xlna

Now differentiate implicitly

1/y = (dx/dy)lna

Note here lna is just a constant, then rearranging we have

dy/dx = ylna

and since y=a^x

dy/dx = a^x(lna)

Answered by James M. Maths tutor

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