In primul rand ai nevoie de tabelul de derivate sa il ai in fata
Il atasez in poza
f(x)=x²-5x+5
Intotdeauna puterea cade in fata si scade cu o unitate
f'(x)=2x-5
f(x)=ln(x²+5x)
Aceasta este o derivata compusa (vezi tabel derivate compuse)
[tex]f'(x)=\frac{(x^2+5x)'}{x^2+5x} =\frac{2x+5}{x^2+5x}[/tex]
[tex]f(x)=\sqrt{9-x^2}[/tex]
este o derivata compusa
[tex]f'(x)=\frac{(9-x^2)'}{2\sqrt{9-x^2} } =-\frac{2x}{2\sqrt{9-x^2} } =-\frac{x}{\sqrt{9-x^2} }[/tex]
[tex]f(x)=2^x[/tex]
derivata simpla
[tex]f'(x)=2^x\times ln2[/tex]
f(x)=ln(1-x)
derivata compusa
[tex]f'(x)=\frac{-1}{1-x}[/tex]
f(x)=(x+9)√x
f'(x)=f'×g+f×g'
f'(x)=(x+9)'√x+(x+9)√x'
[tex]f'(x)=\sqrt{x} +(x+9)\frac{1}{2\sqrt{x} }[/tex]
[tex]f(x)=\frac{x-1}{x+1} \\\\\frac{f'g-fg'}{g^2} \\\\\\f'(x)=\frac{(x-1)'(x+1)-(x-1)(x+1)'}{(x+1)^2} \\\\\\f'(x)=\frac{(x+1)-(x-1)}{(x+1)^2} =\frac{2}{(x+1)^2}[/tex]
Sper ca ti-au fost de folos si ai inteles!
In principiu, urmaresti tabelul cu derivate! daca ai orice intrebare, sunt aici
