MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
nitish kumar Grade:
        

y= tan^-1(secx-tanx), then dy/dx

7 years ago

Answers : (2)

AskIITian Expert Priyasheel - IITD
8 Points
										

d (tan-1(x))/dx = 1/(1+x^2)
To prove the above result,
let y=tan-1(x)
or,tan(y)=x, differentiate w.r.t. x,
sec^2(y).dy/dx=1, or,
dy/dx=cos^2(y) ...(i)
Let p=tan^-1(x), or x=tan(p), so cos(p)=1/(1+x^2)^0.5,
So, cos(p)=cos(tan-1(x))=1/(1+x^2)^0.5.
Therefore, dy/dx=cos^2(y)=cos^2(tan-1(x))=1/(1+x^2)^0.5

In the original question
dy/dx=(sec(x)tan(x)-sec^2(x)) / (1 + (sec(x) - tan(x))^2)
       =-1/2 (after simplification, using 1+tan^2(x)=sec^2(x))

7 years ago
Ramesh V
70 Points
										

derivative of arc tan x is 1/(1+x2)


here dy/dx= ( secx.tanx - sec2x) / (1+sec2x+tan2x-2.secx.tanx)


               = secx(secx-tanx)/2.tanx(secx-tanx)


               =secx/2tanx


               =(cosec x) /2




--


regards


Ramesh

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: Rs. 15,900
  • View Details
Get extra Rs. 3,180 off
USE CODE: CART20
Get extra Rs. 339 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details