Question
Fri January 28, 2011 By: Pooja Dev

# please give the derivative of root of tan root of x using first principal

Sat January 29, 2011
Dear student,

[sin(Ã¢ÂˆÂš(x+h) cos(Ã¢ÂˆÂšx) -sin(Ã¢ÂˆÂšx)cos(Ã¢ÂˆÂšx+h) ] /cos(Ã¢ÂˆÂšx+h)cos(Ã¢ÂˆÂšx) h

sin( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx)
-----------------------
cos(Ã¢ÂˆÂšx+h)cos(Ã¢ÂˆÂšx) h

now lim sin x /x =1 but for this whatever is there in sine should be in
h->0
denominator

so for sin( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) we need

sin( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) * ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) / ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx)

lim sin( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx)
-------------------- * -------------------------------
h->0 ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) cos(Ã¢ÂˆÂšx+h)cos(Ã¢ÂˆÂšx) h

lim sin( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx)
-------------------- =1
h->0 ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx)

so we have

lim ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) )
-------------------- = lim ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) ) 1
h->0 cos(Ã¢ÂˆÂšx+h)cos(Ã¢ÂˆÂšx) h h->0 ----------------------- * ----------------
h cos(Ã¢ÂˆÂšx)cos(Ã¢ÂˆÂšx)

multiplying by conjugate

lim ( Ã¢ÂˆÂš(x+h) -Ã¢ÂˆÂšx) ) ( Ã¢ÂˆÂš(x+h) +Ã¢ÂˆÂšx) ) * 1
----------------------------------------Ã¢Â€Â¦
h->0 h ( Ã¢ÂˆÂš(x+h) + Ã¢ÂˆÂšx) ) cos^2(Ã¢ÂˆÂšx)

lim ( (Ã¢ÂˆÂš(x+h))^2 -(Ã¢ÂˆÂšx)^1 * 1
h->0 ----------------------------------------Ã¢Â€Â¦
h ( Ã¢ÂˆÂš(x+h) + Ã¢ÂˆÂšx) ) cos^2(Ã¢ÂˆÂšx)

lim h
h->0 ----------------------------------------Ã¢Â€Â¦ canceling h and putting h =0
h ( Ã¢ÂˆÂš(x+h) + Ã¢ÂˆÂšx) ) cos^2(Ã¢ÂˆÂšx)

1/(2Ã¢ÂˆÂšx) cos^(Ã¢ÂˆÂšx)

= sec^2 (Ã¢ÂˆÂšx) / ( 2Ã¢ÂˆÂšx)
Hope this helps.
If any queries, please get back to us.
Thanking you
Team
Topperelarning.com
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