EDIT: Sorry, completely missed the existence of a second pageIf i differentiate using your first way Mathstud28
y = (1- √x)²
= (1- √x)(1- √x)
= 1 - 2√x + x
f'(x) = -2(1/2)x^-½ + 1
= -x^-½ + 1
= -1/√x + 1
substituting x->a and y'->? i get this
? = -1/√a + 1
How do i from here show the slope of tha tangent at that √x + √y = 1 at P(a,b) is
Well then do it the second way I showed you
I think we can finally put this to rest.
The thing to remember what im sayin here is that this worksheet i got in school, was given before the class learned implicit differentiation. So That's my problem here how do i prove it equals that without having to even think about implicit differentiation, if i was a currently enrolled calc student in class that didnt know what implicit differentiation meant.
Then what you will have to do is find a way to relate what you got by normal differentiation back to your original function