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### How do you go about differentiating a^x functions?

A key point to remember here is that a^x could mean the base (a) is not “e”-that special number which has a gradient function, dy/dx (the differential), EQUAL to the function itself, y=e^x. When "a" is any real number, you must treat it differently to functions with the base "e".

For the function y=a^x, the rule is simply dy/dx=a^x*lna

(^ means “to the power of”. * means “multiply”)

The proof for this requires an understanding of implicit differentiation (differentiating both x and y terms within an implicit relation).

y=a^x

ln (natural log) both sides of the equation

lny=ln(a^x)

=xlna

Differentiate implicitly

d(lny)/dx=d(xlna)/dx​

(the implicit part means you differentiate with respect to y but multiply by dy/dx after...)

1/y*dy/dx=lna (<----"xlna" differentiates to simply lna because a is just a constant and so lna is also just a constant)

Multiply across by y

dy/dx=ylna

Since y=a^x...

dy/dx=a^x*lna!

N.B. differentiating a^(f(x)) where f(x) is a function of x requires use of the chain rule too (first set the function f(x) to another letter, say u, so you have a^u. Differentiate a^u with the above rule. Differentiate u=f(x) separately, then use chain rule).

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