Question

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

Answer 1

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 

 

 

 

 

 

 

Answer 2

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

 

 

 

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