Răspuns:
Explicație pas cu pas:
g. f(x)=x²lnx
f'(x)=2xlnx+x
f''(x)=2lnx+3
h. f(x)=sin²x
f'(x)=2sinxcos
f''(x)=2cos²x-2sin²x
i. f(x)=cos³x
f'(x)=-3cos²xsinx
f''(x)=-3(-2cosxsinx-cos³x)=6cosxsinx+3cos³x
j. f(x)=xsinx+cosx
f'(x)=sinx-xcos-sinx=-xcosx
f''(x)=-cosx+xsinx
k. f(x)=x²√x
[tex]f'(x)=2x\sqrt{x} +x^{2} \cdot \frac{1}{2\sqrt{x} } =2x\sqrt{x} +\frac{x\sqrt{x} }{2} =\frac{5x\sqrt{x} }{2}[/tex]
[tex]f''(x)=\frac{5}{2} (\sqrt{x} +x\cdot \frac{1}{2\sqrt{x} } )=\frac{15\sqrt{x} }{4}[/tex]
l. f(x)=xtgx
[tex]f'(x)=tgx+\frac{x}{cos^{2}x }[/tex]
[tex]f''(x)=\frac{1}{cos^{2}x } +\frac{cos^{2}x+2xcoxsinx }{cos^{4}x } =\frac{2cosx+2xsinx}{cos^{3}x }[/tex]
m. f(x)=[tex]\frac{x-1}{x+2}[/tex]
[tex]f'(x)=\frac{x+2-x+1}{(x+2)^{2} } =\frac{3}{(x+2)^{2}} \\f''(x)=\frac{-(x+2)^{2}-6(x+2) }{(x+2)^{4}} =\frac{-x^{2} -10x-16}{(x+2)^{4}}=- \frac{(x+2)(x+8)}{(x+2)^{4}} =-\frac{x+8}{(x+2)^{3}}[/tex]
n. [tex]f(x)=\frac{x}{x^{2} +1}[/tex]
[tex]f'(x)=\frac{x^{2} +1-x(2x)}{(x^{2} +1)^{2} } =\frac{-x^{2} +1}{(x^{2} +1)^{2} }[/tex]
[tex]f''(x)=\frac{-2x(x^{2} +1)^{2}-2(1-x^{2} )(x^{2} +1)2x }{(x^{2} +1)^{4} } =\frac{-2x[(x^{2} +1)+2(1-x^{2} )]}{(x^{2} +1)^{3}} =\frac{-2x(-x^{2} +3)}{(x^{2} +1)^{3}}[/tex]
sa mai verifici si tu calculele inca odata. le-am facut direct de pe laptop si e greu sa urmaresc tot
