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differentiation help

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lhl_anecar:
hi... i need some help with these questions. :D

Find the derivative of the following,which rule should i use, and how do i go about it?
Q1  sec x x cosec 2x

Q2  (tan^-1) x + ln((x^2)+1)

Q3 (sin^-1)x + squareroot of (1-x^2)

Thanks!!!

Chingoo:
[o]Remember to always convert cosec, sec and cot into their parent trigonometric ratios.
1. secx=1/cosx
2. cosecx=1/sinx
3. cotx=1/tanx[o]Also keep in mind identities you may need at times for substitution:
4. sin2x + cos2x = 1
5. tanx = sinx/cosx
6. 1 + cot2x = cosec2x
7. tan22 + 1 = sec2x
8. Double angle identities:
 (a) sin2A = 2sinAcosA
 (b) cos2A = cos2A + sin2A = 2cos2A - 1 = 1 - 2sin2A
 (c) tan2A = 2tanA/(1 - tan2A)[o]The derivative of these parent trigonometric ratios can be found out by the concepts:
9. d(cosx)/dx = -sinx d(x)/dx
10. d(sinx)/dx = cosx d(x)/dx[o]Also, the general functions of derivation:
11. y = u.v ; dy/dx = u.d(v)/dx + v.d(u)/dx (Order of factors does not matter i.e. v.d(u)/dx + u.d(v)/dx  is also correct)
12. y = u/v ; dy/dx = [v.d(u)/dx - u.d(v)/dx]/v2 {Order of factors must not be changed)
13. y = axn ; dy/dx = anxn-1.d(x)/dx
14. y = ex ; dy/dx = ex. d(x)/dx
15. y = lnx ; dy/dx = (1/x).d(x)/dx
Now, to your questions. I'll attempt one and then you try the other two. :)

Q1.
(secx)(cosec2x)
dy/dx = secx. d(cosec2x)/dx + cosec2x. d(secx)/dx {see 11}
dy/dx = (1/sinx).d(1/sin2x)/dx + (1/sin2x). d(1/cosx)/dx  {see 1, 2 & 3)
dy/dx = (1/sinx).([sin2x.d(1)/dx – d(sin2x)/dx]/sin22x) + (1/sin2x). ([cosx.d(1)/dx – d(cosx)/dx]/cos2x) {see 12}
dy/dx = (1/sinx).(-2cos2x/sin22x) + (1/sin2x). (sinx/cos2x) {see 9 & 11}
dy/dx = sinx/( cos2x)(sin2x) – 2cosx/( sin22x)(sinx)

Working doesn’t necessarily have to stop here—it depends on the requirement of the question (sometimes it wants you to prove or show something) where you can stop altering the format of the equation.[/list]

lhl_anecar:

--- Quote from: Chingoo on October 31, 2010, 01:01:56 am ---[o]Remember to always convert cosec, sec and cot into their parent trigonometric ratios.
1. secx=1/cosx
2. cosecx=1/sinx
3. cotx=1/tanx[o]Also keep in mind identities you may need at times for substitution:
4. sin2x + cos2x = 1
5. tanx = sinx/cosx
6. 1 + cot2x = cosec2x
7. tan22 + 1 = sec2x
8. Double angle identities:
 (a) sin2A = 2sinAcosA
 (b) cos2A = cos2A + sin2A = 2cos2A - 1 = 1 - 2sin2A
 (c) tan2A = 2tanA/(1 - tan2A)[o]The derivative of these parent trigonometric ratios can be found out by the concepts:
9. d(cosx)/dx = -sinx d(x)/dx
10. d(sinx)/dx = cosx d(x)/dx[o]Also, the general functions of derivation:
11. y = u.v ; dy/dx = u.d(v)/dx + v.d(u)/dx (Order of factors does not matter i.e. v.d(u)/dx + u.d(v)/dx  is also correct)
12. y = u/v ; dy/dx = [v.d(u)/dx - u.d(v)/dx]/v2 {Order of factors must not be changed)
13. y = axn ; dy/dx = anxn-1.d(x)/dx
14. y = ex ; dy/dx = ex. d(x)/dx
15. y = lnx ; dy/dx = (1/x).d(x)/dx
Now, to your questions. I'll attempt one and then you try the other two. :)

Q1.
(secx)(cosec2x)
dy/dx = secx. d(cosec2x)/dx + cosec2x. d(secx)/dx {see 11}
dy/dx = (1/sinx).d(1/sin2x)/dx + (1/sin2x). d(1/cosx)/dx  {see 1, 2 & 3)
dy/dx = (1/sinx).([sin2x.d(1)/dx – d(sin2x)/dx]/sin22x) + (1/sin2x). ([cosx.d(1)/dx – d(cosx)/dx]/cos2x) {see 12}
dy/dx = (1/sinx).(-2cos2x/sin22x) + (1/sin2x). (sinx/cos2x) {see 9 & 11}
dy/dx = sinx/( cos2x)(sin2x) – 2cosx/( sin22x)(sinx)

Working doesn’t necessarily have to stop here—it depends on the requirement of the question (sometimes it wants you to prove or show something) where you can stop altering the format of the equation.[/list]

--- End quote ---

Thanks a lot!!! But i am abit confuse, shouldn't the part in red be (1/cos x ) ???

Deadly_king:

--- Quote from: lhl_anecar on October 31, 2010, 03:05:51 pm ---Thanks a lot!!! But i am abit confuse, shouldn't the part in red be (1/cos x ) ???

--- End quote ---

Yupz.............you are right.

sec x = 1/cos x

I guess it was a typing error. Sorry for the confusion. ;)

Chingoo:
Yes, sorry about that ^_^ Thank you Deadly_King for clearing that up :D

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