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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(â(x+h) cos(âx) -sin(âx)cos(âx+h) ] /cos(âx+h)cos(âx) h

sin( â(x+h) -âx)
-----------------------
cos(âx+h)cos(âx) h

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

so for sin( â(x+h) -âx) we need

sin( â(x+h) -âx) * ( â(x+h) -âx) / ( â(x+h) -âx)

lim sin( â(x+h) -âx) ( â(x+h) -âx)
-------------------- * -------------------------------
h->0 ( â(x+h) -âx) cos(âx+h)cos(âx) h

lim sin( â(x+h) -âx)
-------------------- =1
h->0 ( â(x+h) -âx)

so we have

lim ( â(x+h) -âx) )
-------------------- = lim ( â(x+h) -âx) ) 1
h->0 cos(âx+h)cos(âx) h h->0 ----------------------- * ----------------
h cos(âx)cos(âx)

multiplying by conjugate

lim ( â(x+h) -âx) ) ( â(x+h) +âx) ) * 1
----------------------------------------â¦
h->0 h ( â(x+h) + âx) ) cos^2(âx)

lim ( (â(x+h))^2 -(âx)^1 * 1
h->0 ----------------------------------------â¦
h ( â(x+h) + âx) ) cos^2(âx)

lim h
h->0 ----------------------------------------â¦ canceling h and putting h =0
h ( â(x+h) + âx) ) cos^2(âx)

1/(2âx) cos^(âx)

= sec^2 (âx) / ( 2âx)
Hope this helps.
If any queries, please get back to us.
Thanking you
Team
Topperelarning.com
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