- 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 = 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]/sin
22x) + (1/sin2x). ([cosx.d(1)/dx – d(cosx)/dx]/cos
2x)
{see 12}dy/dx = (1/sinx).(-2cos2x/sin
22x) + (1/sin2x). (sinx/cos
2x)
{see 9 & 11}dy/dx = sinx/( cos
2x)(sin2x) – 2cosx/( sin
22x)(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]