Differentiate y= 2^x

Initially this looks unlike all the other differentiation questions and seems unsolvable. However the expression 2^x can be rewritten in an equivalent form that will allow us to use the differentiation rules we already know. We know that e^(ln(x)) is the same as x, consequently e^(ln(2^x)) is 2^x. We know how to differentiate e to the power of a function of x by using the chain rule. If y=e^u, where u= ln(2^x), (this can be rewritten as 2lnx) then we have dy/du= e^u and du/dx= ln2. Multiplying these together to get dy/dx= ln2e^u. The u has to be converted back to its x form, (u=ln(2^x)), dy/dx= ln22^x. As long as the first step is remembered the rest is just applying the differentiation rules we already know.