I’m embarrassed to admit I’m getting my masters degree in a math related subject and I still don’t get this!
I know how to do them, but the way you compute them is almost suspiciously simple. What’s the logic behind converting the exponent to a constant? How does that determine the slope?
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The source of the “2” in the derivative formula (x^(2))’ = 2x is the middle term in the algebraic identity
(a+b)^2 = a^2 + 2ab + b^2
because in the definition of the derivative of x^(2), we compute (for nonzero h)
((x + h)^2 – x^(2))/h = (x^2 + 2xh + h^2 – x^(2))/h = 2x + h
and the limit of that as h tends to 0 is 2x. So the “2” in the derivative formula 2x comes from the second term in the expansion of (x+h)^2 that you saw in algebra. Similarly, the 3 in the formula (x^(3))’ = 3x^2 comes from the second term in the cubic expansion
(x+h)^3 = x^3 + 3x^(2)h + 3xh^2 + h^3
after feeding the right side into the limit definition of the derivative of x^(3).
It’s pretty important that these exponents, 2 in x^2 and 3 in x^(3), are *constant* while the base is the variable x. If you swapped those roles and made the base constant and the exponent x, then everything is totally different: you’re now dealing with exponential functions 2^x and 3^x rather than polynomials x^2 and x^(3). Exponential functions have different properties and different graphs compared to polynomials.
It might be natural to guess that 2^x has derivative x2^(x-1), but that’s wrong. Think about what’s happening at x = 0: the graph of y = 2^x when it passes through the y-axis x = 0 is going up, so (2^(x))’ at x = 0 is positive, in fact (2^(x))’ at all x is positive, but x2^(x-1) vanishes when x = 0 and is negative when x < 0, so the formula x2^(x-1) is not at all like (2^(x))’.
The correct formula for (2^(x))’ is (2^(x))ln(2), and (3^(x))’ = (3^(x))’ln(3): we need logarithms to describe derivatives of exponential functions, which is far from obvious when you first see these things. A reason that 2^(x) has such a different derivative formula than x^2 is the different mathematical properties of 2^x compared to x^(2). For instance, these functions have completely different effects on sums: (a+b)^2 = a^2 + 2ab + b^2 while 2^(a+b) = 2^(a)2^(b).
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