IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: lhl_anecar on October 30, 2010, 07:44:30 pm

Title: differentiation help
Post by: lhl_anecar on October 30, 2010, 07:44:30 pm

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!!!

Title: Re: differentiation help
Post by: Chingoo on October 31, 2010, 01:01:56 am

• 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

• 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)

• 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

• 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 = anx^n-1.d(x)/dx
14. y = e^x ; dy/dx = e^x. 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]

Title: Re: differentiation help
Post by: lhl_anecar on October 31, 2010, 03:05:51 pm

Quote from: Chingoo on October 31, 2010, 01:01:56 am

• 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

• 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)

• 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

• 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 = anx^n-1.d(x)/dx
14. y = e^x ; dy/dx = e^x. 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]

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

Title: Re: differentiation help
Post by: Deadly_king on November 01, 2010, 05:28:48 pm

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 ) ???

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

sec x = 1/cos x

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

Title: Re: differentiation help
Post by: Chingoo on November 02, 2010, 09:47:27 am

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

Title: Re: differentiation help
Post by: Deadly_king on November 02, 2010, 10:21:49 am

Quote from: Chingoo on November 02, 2010, 09:47:27 am

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

Mention not Miss Chingoo ;)

Was just doing my job. :D