Why do we derive 1+(x-1)^2 to 2(x-1) don't we have to derive first the outer function and subsequently the inner one? So that the result would be 2?

Question

anonymous · Answer

Never mind.

jrmorganjr · Answer

I guess you've reasoned it through, but order of operations is signs first. The other way to look at is "x-1" is really the same as "x", or can be written as y, and you follow the pattern for y^2, then put (x-1) back in.

jkristia · Answer

>>The other way to look at is "x-1" is really the same as "x
I dont think this is correct. Think of
f(x) = (2x+3)^2.
f'(x) is not 2(2x+3), but instead (by the product rule or by multiplying out (2x+3) )

f'(x) = 2(2x+3)+2(2x+3) = 8x+12

It just happens to work for (x-1)^2 since (x-1)^2 = x^2-2x-2 and the derivative is 2x-2, or 2(x-1)

jrmorganjr · Answer

Sorry, sloppy notation. For your example, f(x)= (2x+3)^2 becomes f(x) = y^2 with y=2x+3. f'(x) = 2y dy/dx = 2 (2x+3) 2 = 8x+12, just as you did by multiplying it out. I like doing it this way both because it's simple and clear but mainly because higher powers become progressively difficult to mulitply out, not to mention inelegant.