2)y= (In x)^4= [tex] y^{,} =4(lnx)^3( \frac{dx}{x})[/tex]
4)f(x)=[tex] In(\frac{x}{x^{2} +1})dx [/tex] >>>> Es parecida a la "3)"
=[tex] f^{,}(x)= \frac{ \frac{1*(x^{2} +1)-x(2x)}{(x^{2} +1)^2}}{ \frac{x}{x^{2} +1}}dx[/tex]
=[tex] \frac{ \frac{x^{2} +1-2x^2}{(x^{2} +1)^2}}{ \frac{x}{x^{2} +1}}dx[/tex]
=[tex] \frac{ \frac{-x^2+1}{(x^{2} +1)^2}}{ \frac{x}{x^{2} +1}}dx[/tex]
=[tex]\frac{(-x^2+1)(x^2+1)}{x(x^{2} +1)^2}dx[/tex] Aplicando "Doble C"
=[tex]\frac{-x^2+1}{x(x^{2} +1)}}dx[/tex] Simplificando
6)y= In (In x^2)=
[tex] y^{,} =\frac{1}{ln(x^2)}*\frac{1}{x^2}*2xdx[/tex]
[tex]\frac{2x}{x^2ln(x^2)}=\frac{2}{xln(x^2)}dx[/tex] Simplificando
8)g(x)=3(4- 9x)^4 Es parecida a la "7"
[tex] g^{,}(x) =12(4-9x)^3*(-9)=108(4-9x)^3*dx[/tex]
10)y=∛(9x^2+4)
[tex] y^{,} =\frac{1}{3* \sqrt[3]{(9x^2+2)^2} }*(18x)dx[/tex]
=[tex]\frac{18x}{3* \sqrt[3]{(9x^2+2)^2} }dx[/tex]
=[tex]\frac{6x}{ \sqrt[3]{(9x^2+2)^2} }dx[/tex]
12)y= 1/(x - 2)=> [tex] y^{,}= \frac{-1}{(x-2)^2}dx[/tex]
14)g(x)=3 arccos x/2 (Voy a asumir que el argumento es x/2 AUNQUE no hay parentesis
[tex] g^{,}(x) = 3*(-\frac{1}{ \sqrt{1-x^2}})*\frac{1}{2}dx[/tex]
[tex] =-\frac{3}{2 \sqrt{1-x^2}}dx[/tex]
16)g(x)= arcsen 3x/x =g(x)= arcsen 3
(mal planteada. No hay ningún aungulo que produzca senx=3)
17)h(t)=sen(arccos t)
[tex] h^{,}(t) =Cos(ArcCost) (-\frac{dt}{ \sqrt{1-x^2}})[/tex]
18)y= x arccos x -√(1 -x^2)
[tex] y^{,} =ArcCosx+(-\frac{x}{ \sqrt{1-x^2}})dx-\frac{-2x}{ 2\sqrt{1-x^2}}dx[/tex]
[tex]=ArcCosx+(-\frac{xdx}{ \sqrt{1-x^2}})+\frac{xdx}{ \sqrt{1-x^2}}[/tex]
[tex]=ArcCosx[/tex] >>>>(Como los dos últimos términos son de signos opuestos, se anulan)
20)y=1/2[x√4-x^2 + 4 arcsen (x/2)]
[tex]y^{,} =\frac{1}{2} (\sqrt{4-x^2}+x*(\frac{1}{ 2\sqrt{4-x^2}})*-2xdx+4*(\frac{1}{\sqrt{1- (\frac{x}{2})^2}})\frac{1}{2}dx[/tex]
[tex]=\frac{1}{2} (\sqrt{4-x^2}-(\frac{x^2dx}{ \sqrt{4-x^2}})+\frac{2dx}{\sqrt{1- \frac{x^2}{4}}})[/tex]
22)6(2x-7)^2 >>>12(2x-7)*2=24(2x-7)=48x-168
Espero te sea de utilidad. No olvides xfav calificar nuestras respuestas. Exito
