IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

gayathri

how should we differetiate modulus functions.
eg
y=1/|sinx|
dy/dx=?

ritu_007

hi!

d/dx IxI=IxI/x

y=IcosecxI

dy/dx= IcosecxI/cosecx * -cosecx.cotx

= -IcosecxI.cotx.

ramyadiamond

u need to break up the function before differenciating, as below,

y= { 1/sinx if x lies between 0 and pi}

and

{-1/sinx if x lies between pi and 2pi}

so, dy/dx = { -cotx.cosecx if x lies between 0 and pi}

and

{cotx.cosecx if x lies between pi and 2pi}

so for every mod function, first make cases acc. to domain and only then differenciate it. Here, i haven't taken the general value for x, but u modify it for any general value.

hope u got it!!

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iitkgp_bipin

f(x) = 1/|sinx|

Break the function :

f(x) = 1/sinx x

(2n

,(2n+1)

)

= -1/sinx x

((2n-1)

,2n

)

At x=2n

function is neither continuous nor differentiable.

And in the above intervals they are differentiable. Simply differentiate the function in these intervals.

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