найти производную: у=sin(arctg^4(4x^3)) y=tg(arccos^3(3x^2))

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найти производную: у=sin(arctg^4(4x^3)) y=tg(arccos^3(3x^2))


Математика (22 баллов) | 37 просмотров
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у' =(sin(arctg^4(4x^3)))' = cos(arctg^4(4x^3))*(arctg^4(4x^3))' = 

 = cos(arctg^4(4x^3))*(4arctg^3(4x^3))*(arctg(4x^3))' = 

= cos(arctg^4(4x^3))*(4arctg^3(4x^3))*(1/(1+(4x^3)^2)* (4x^3)' =

=  cos(arctg^4(4x^3))*(4arctg^3(4x^3))*(1/(1+(4x^6))* 12x^2 

 

 

 

 y' = (tg(arccos^3(3x^2)))' = 1/cos^2(arccos^3(3x^2))*(arccos^3(3x^2))'  =

=  1/cos^2(arccos^3(3x^2))*3arccos^2(3x^2)*(arccos(3x^2)' =

=  1/cos^2(arccos^3(3x^2))*3arccos^2(3x^2)*(-1/корень(1-(3x^2)^2))*(3x^2)' =

=  1/cos^2(arccos^3(3x^2))*3arccos^2(3x^2)*(-1/корень(1-(3x^4))*6x 

 

 

 

 

 

 
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y=sin(arctg^4(4x^3))\\\\y'=cos(arctg^4(4x^3))\cdot4arctg^3(4x^3)\cdot\frac{1}{1+16x^6}\cdot12x^2

 

 

 

\bf y=tg(arccos^3(3x^2))\\\\y'=\frac{1}{cos^2(arccos^3(3x^2))}\cdot3arccos^2(3x^2)\cdot(-\frac{1}{\sqrt{1-9x^4}})\cdot6x

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