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a fost răspuns

Ajutor la urm derivate



f(x)= x la a 3 + lnx+ e la puterea x



f(x)=x la puterea x +x+ sinx



f(x)= 1 supra x la puterea 3 + 1 supra x la puterea 2 + 1 supra x +3


f(x)= log 2 la x + log 3 la x





Răspuns :

[tex]1) f^{'}(x)= ( x^{3}+ln x+ e^{x} )^{'} = ( x^{3} )^{'} + (ln x)^{'} + (e^{x})^{'} =3 x^{2} + \frac{1}{x}+ e^{x} \\ 2) f^{'}(x)= ( x^{x}+x+sinx )^{'}=( x^{x} )^{'}+x^{'}+(sinx){'}=x*x^{x-1}+1+\\+cos x=x^{1+x-1}+1+cosx=x^{x}+1+cosx\\ 3) f^{'}(x)=( \frac{1}{ x^{3} }+ \frac{1}{ x^{2} } + \frac{1}{x+3} )^{'}=( x^{-3} )^{'}+( x^{-2} )^{'}+(x-3)^{'}=-3 x^{-4}-\\- 2x^{-3}-(x+3)*x^{'}= -3 x^{-4}- 2x^{-3}-x-3\\ [/tex]
[tex]4) f^{'}(x)=( log_{2}x+log_{3} x)^{'}=(log_{2}x)^{'}+(log_{3}x)^{'}=\\= \frac{1}{xln2}+ \frac{1}{xln3}= \frac{1}{x}( \frac{1}{ln2}+ \frac{1}{ln3}) [/tex]