y = arc ctg 2/x+ arc tg x/2

Tentukanlah turunan pertama dari y = arc ctg 2/x+ arc tg x/2 !

Jawab:

Karena:

d/dx ( arc ctg u)= (- 1)/(1+ u^2 ) (du/dx)

Maka :

d/dx ( arc ctg 2/x)= (- 1)/(1+ (2/x)^2 ) (d(2/x)/dx)

d/dx ( arc ctg 2/x)= (- 1)/(1+ 4/x^2 ) ((-2)/x^2 )

d/dx ( arc ctg 2/x)= (-1)/((x^2+4)/x^2 ) ((-2)/x^2 )

d/dx ( arc ctg 2/x)=((-x^2)/(x^2+ 4)) ((-2)/x^2 )

d/dx ( arc ctg 2/x)=(2/(x^2+ 4))

Karena :

d/dx ( arc tg u) = 1/(1+ u^2 ) (du/dx)

Maka:

d/dx ( arc tg x/2)= 1/(1+ (x/2)^2 ) (d(x/2)/dx)

d/dx ( arc tg x/2)= 1/(1+ x^2/4) (2/4)

d/dx ( arc tg x/2)=(4/(4+ x^2 )) (2/4)

d/dx ( arc tg x/2)=(2/(4+ x^2 ))

Maka turunan dari y = arc ctg 2/x+ arc tg x/2 adalah

dy/dx= d/dx ( arc ctg 2/x)+ d/dx ( arc tg x/2)

dy/dx= (2/(x^2+ 4))+ (2/(4+ x^2 ))

dy/dx= ((2+2)/(x^2+ 4))

dy/dx= (4/(x^2+ 4))

 

for a more clear author please click the link below

Tentukanlah turunan pertama dari y = arc ctg 2/x+ arc tg x/2