Răspuns :
[tex]\it 2ln(x^2-9)=ln(x^2-9)^2[/tex]
[tex]\it (ln \ u)'=\dfrac{u'}{u}[/tex]
[tex]\it f'(x)=\dfrac{1}{x-3}-\dfrac{2(x^2-9)\cdot2x}{(x^2-9)^2}=\dfrac{1}{x-3}-\dfrac{4x}{x^2-9}=\dfrac{x+3-4x}{x^2-9}=\dfrac{-3x+3}{x^2-9}[/tex]
Răspuns:
Se aplica formula ln `u=u`/u
ln`(x-3)=(x-3) `/(x-3)=1/(x-3)
ln `(x²-9)=(x²)`/(x²-9)=2x/(x²-9)
f `(x)=1/(x-3)-4x/X²-9)=
1/(x-3)-4x/(x-3)(x+3)=
(x+3-4x)/(x-3)(x+3)=
(-3x+3)/(x²-9)
Explicație pas cu pas: