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

Asked by pooja dev 28th January 2011, 12:00 AM
Dear student,

[sin(√(x+h) cos(√x) -sin(√x)cos(√x+h) ] /cos(√x+h)cos(√x) h

sin( √(x+h) -√x)
-----------------------
cos(√x+h)cos(√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( √(x+h) -√x) we need

sin( √(x+h) -√x) * ( √(x+h) -√x) / ( √(x+h) -√x)

lim sin( √(x+h) -√x) ( √(x+h) -√x)
-------------------- * -------------------------------
h->0 ( √(x+h) -√x) cos(√x+h)cos(√x) h

lim sin( √(x+h) -√x)
-------------------- =1
h->0 ( √(x+h) -√x)

so we have

lim ( √(x+h) -√x) )
-------------------- = lim ( √(x+h) -√x) ) 1
h->0 cos(√x+h)cos(√x) h h->0 ----------------------- * ----------------
h cos(√x)cos(√x)

multiplying by conjugate

lim ( √(x+h) -√x) ) ( √(x+h) +√x) ) * 1
----------------------------------------…
h->0 h ( √(x+h) + √x) ) cos^2(√x)

lim ( (√(x+h))^2 -(√x)^1 * 1
h->0 ----------------------------------------…
h ( √(x+h) + √x) ) cos^2(√x)

lim h
h->0 ----------------------------------------… canceling h and putting h =0
h ( √(x+h) + √x) ) cos^2(√x)

1/(2√x) cos^(√x)

= sec^2 (√x) / ( 2√x)
Hope this helps.
If any queries, please get back to us.
Thanking you
Team
Topperelarning.com
Answered by Expert 29th January 2011, 3:42 PM
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