sign up I login
 advanced
refer a friend - earn nickels!!

Ask & Discuss Questions with Community & Experts

Moderation Team
 90 chars left    advanced
Ask iit jee aieee pet cbse icse state board experts Expert Question: Logarithmic Differentiation Problem
Forum Index -> Differential Calculus like the article? email it to a friend.  
Author Message
Ashish (95)

Scorching goIITian

Olaaa!! Perrrfect answer. 15  [25 rates]

Ashish's Avatar

total posts: 215    
offline Offline
Hello everyone!
 
If y = (sinx)cosx  + (cosx)sinx, prove that
dy/dx = (sinx)cosx . [cotx cosx ? sinx (log sinx)] + (cosx)sinx. [cosx (log cosx) ? sinx tanx]

(CBSE 2004C)
 
The answer that I got was
dy/dx =  [(sinx)cosx. (cosx)sinx] [cotx cosx ? sinx (logsinx) + cosx (log cosx) ? sinx tanx]
 
Kindly explain why I am getting it different. I?ll be grateful to you.
 
Thanks!


With regards

Ashish
 

'Your eyes show the strength of your soul.'
    
Ashish (95)

Scorching goIITian

Olaaa!! Perrrfect answer. 15  [25 rates]

Ashish's Avatar

total posts: 215    
offline Offline
That question mark (?) was actually meant to be the dot (.) for multiplication. I don't know what the problem is but every time it occurs. Any way out?

'Your eyes show the strength of your soul.'
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
smartk (26)

Cool goIITian

Olaaa!! Perrrfect answer. 4  [7 rates]

smartk's Avatar

total posts: 35    
offline Offline
firstly d correct ans is
[sinx^cosx{cotxcosecx-sinxlogsinx} +cosx^sinx{cosxlogcosx+tanxsecx}]
 
take p= sinx^cosx then differentiate it u will get d first expression and then q= cosx^sinx . and add the outcome.
 this reply: 5 points  (with Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
You have to be logged on to rate
  
rvinitra (24)

New kid on the Block

Olaaa!! Perrrfect answer. 4  [6 rates]

rvinitra's Avatar

total posts: 19    
offline Offline
The Question is:
 If y = (sinx)cosx  + (cosx)sinx, prove that
dy/dx = (sinx)cox.[cotx.cosx - sinx(log sinx)] + (cosx)sinx.[cosx (log cosx) - sinx.tanx]

The answer is:
Consider u = (sinx)cosx  and v = (cosx)sinx
Hence y = u + v
Differentiating with respect to x, we get
dy/dx = du/dx + dv/dx ........(1)

Now consider u = (sinx)cosx
Taking log on both sides, we get
log(u) = log(sinxcosx)
log(u) = cosx.log(sinx)
Differentiating with respect to x, we get
1/u . du/dx = -sinx.log(sinx) + (cosx/sinx).cosx
du/dx = u.[cotx.cosx - sinx.log(sinx)]
du/dx = (sinx)cosx[cotx.cosx - sinx.log(sinx)].......(2)

Now consider v = (cosx)sinx
Taking log on both sides, we get
log(v) = log(cosxsinx)
log(u) = sinx.log(cosx)
Differentiating with respect to x, we get
1/v . dv/dx = cosx.log(cosx) + (sinx/cosx)(-sinx)
dv/dx = v.[cox.log(cosx) - tanx.sinx]
dv/dx = (cosx)sinx[cox.log(cosx) - tanx.sinx]......(3)

Now from (1), we get dy/dx = (2) + (3)
Hence
dy/dx = (sinx)cosx[cotx.cosx - sinx.log(sinx)] + (cosx)sinx[cox.log(cosx) - tanx.sinx]

Hence proved.
 this reply: 12 points  (with Olaaa!! Perrrfect answer.   in 3 votes )   [?]
 
You have to be logged on to rate
  
rvinitra (24)

New kid on the Block

Olaaa!! Perrrfect answer. 4  [6 rates]

rvinitra's Avatar

total posts: 19    
offline Offline
I hope u got it. If not please come forward and ask any doubt u hav!!
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
Ashish (95)

Scorching goIITian

Olaaa!! Perrrfect answer. 15  [25 rates]

Ashish's Avatar

total posts: 215    
offline Offline
Hello rvinitra!

Perfect answer! Alas! I have already accepted an answer. You deserve your first salute!

Thanks!

'Your eyes show the strength of your soul.'
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
rvinitra (24)

New kid on the Block

Olaaa!! Perrrfect answer. 4  [6 rates]

rvinitra's Avatar

total posts: 19    
offline Offline
thank u
 this reply: 0 points  (with Olaaa!! Perrrfect answer.   in 0 votes )   [?]
 
You have to be logged on to rate
  
 
Forum Index -> Differential Calculus
Go to:   

Top Offers for goIITians
Correspondence Courses
Brilliant Tutorials
Narayana Institute
Aakash Institute
Classroom/Crash Courses
Narayana - Kota , Delhi , Others
Brilliant Tutorials - Class , Crash
Aakash Institute - Medical , Engg
Online Test Series
Brilliant Tutorials
Narayana Institute
Aakash Institute
Mahesh Tutorials
AMITY      Sri Chaitanya