Question

Va rog sa ma ajutati cu exercitiul din imagine.

Answer 1

s e stio (???) ca (f/g)'=(f'g-fg')/g²

atunci

f'(x) =(2x+6)(x-2)-(x²+6x)/(x-2)²=(2x²+2x-12-x²-6x)/(x-2)²=(x²-6x+2x-12)/(²x-2)=

=(x(x-6)+2(x-6))/(x-2)²= (x-6)(x+2)/(x-2)².C.C.T.D.

Answer 2

Folosim relatia: $(\frac{f}{g})^{'}=\frac{f^{'}g-fg^{'}}{g^{2}}$

In cazul din exemplu:

$f^{'}=\frac{(x^{2}+6x)^{'}(x-2)-(x^{2}+6x)(x-2)^{'}}{(x-2)^{2}}=\frac{(2x+6)(x-2)-(x^{2}+6x)}{(x-2)^{2}}=\frac{x^{2}-4x-12}{(x-2)^{2}}=\frac{x^{2}-6x+2x-12}{(x-2)^{2}}=\\\frac{(x-6)(x+2)}{(x-2)^{2}}$