IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

Prajju

y = e^{1/2 log(1+tan^2)}then find dy/dx......

Ans ==> sec xtan x

paddy.dude

its easy
first take log on both sides
u will get
log y= 1/2 log 1+ tan^2 x
take the 1/2 inside and we know 1+tan^x =sec^2x

so it becomes log y= log sec x

ie y=sec x
and now this is a std differential

gkandoi

given : Y=e^{1/2LOG(1+TANX)}

TAKE LOG ON BOTH SIDES.......U GT

LOG Y =I/2 LOG(SEC^2 X)

DIFF BOTH SIDES

1/Y DY/DX = 1/2 2.SECX.SECX.TANX/SEC^2 X

DY/DX = e^{1/2 LOG(SEC^2 X)}TANX

TAKE THE SQ. OF SECX OUT OF LOG....1/2 AND 2 CANCELS.....THN APPLY THE BASE CHANGE PROPERTY OF LOG....i.e TAKE E IN LOG AND SECX DOWN.....

HERE'S UR ANSWER....

REPLY IF U DINT UNDERSTOOOD NY STEP...

aditi_g

write 1+tan^2 x as sec^2 x
e^1/2 log sec^2 x = e^log secx
therefore
y = sec x
dy/dx = sec x tanx

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y = e1/2 log(1+tan^2) then find dy/dx......

y = e^{1/2 log(1+tan^2)}then find dy/dx......