How do I differentiate 3^x using first principles ?

{lim h-> 0} [3^(x+h) - 3^x] / h

={lim h-> 0} (3^x)[3^h - 1] / h

The only way to evaluate this is by using L'Hopital's rule but this is a circular proof since you would need to differentiate 3^h. Instead I would write 3^x as

e^(ln3^x) = e^(x * ln3)

And then differentiate it to ln3 * 3^x.

When x=1 the answer is just 3 * ln3.

f(x) = 3x^2 -5 then f(x+h) = { 3(x+h)^2 - 5 [ d/dx 5 =0] d/dx 3x^2 --d/dx 5 = d/dx 3x^2 -0 now d/dx 3x^2 =limh-- >0 3{ ( x+h)^2 -x^2}/h x isn't equivalent to 0 so limh---->0 3x^2[ {a million+h/x)^2 --a million]/h x isn't equivalent to 0 also h has a tendency to 0. so we would think hto be numerically smller than x i. e. h/x to be smaller than solidarity .we may be able to hence advance (a million+h/x)^2 by biomial theorem and write as limh---->0 3 x^2 [a million+2h/x +(h/x)^2 -a million]/h so limh-->0 3x^2 [2/x +h/x^2]= lim h--->0 3x^2 [ 2/x + (0)] d/dx [3x^2 -5] =3x^2 .2/x =6x ans

x3^(x-1)

x3^(x-1)