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^xlna(^ 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).Start with the functiony=a^xln (natural log) both sides of the equationlny=ln(a^x)    =xlnaDifferentiate implicitlyd(lny)/dx=d(xlna)/dx​(the implicit part means you differentiate with respect to y but multiply by dy/dx after...)1/ydy/dx=lna (<----"xlna" differentiates to simply lna because a is just a constant and so lna is also just a constant)Multiply across by ydy/dx=ylnaSince 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).