lesson 11 derivative of trigonometric functions
TRANSCRIPT
DIFFERENTIATION OF TRIGONOMETRIC
FUNCTIONS
OBJECTIVES:• differentiate functions involving trigonometric
functions;• solve problems involving differentiation of
trigonometric functions; and• apply trigonometric identities to simplify the
functions.
TRANSCENDENTAL FUNCTIONS
Kinds of transcendental functions:1.logarithmic and exponential functions2.trigonometric and inverse trigonometric functions3.hyperbolic and inverse hyperbolic functions Note:Each pair of functions above is an inverse to each other.
The TRIGONOMETRIC FUNCTIONS.
xtan1
xsinxcosxcot .4
xcot1
xcosxsinxtan .3
x cos1 x sec
xsec1xcos 2.
x sin1x csc
xcsc1xsin 1.
Identities ciprocalRe .AIdentities ricTrigonomet
:callRe
ytanxtan1
ytanxtanyxtan .3
ysinxsinycosxcosyxcos 2. ysinxcosycosxsinyxsin 1.
Angles Two of Difference and Sum.B
xtan1xtan2x2tan .3
1xcos2
xsin21
xsinxcosx2cos 2.
2sinxcosx x2sin 1. Formulas Angle Double .C
2
2
2
22
xcscxcot1 .3
xsecxtan1 .2
1xcosxsin .1
Identities Squared .D
22
22
22
DIFFERENTIATION FORMULADerivative of Trigonometric FunctionFor the differentiation formulas of the trigonometric functions, all you need to know is the differentiation formulas of sin u and cos u. Using these formulas and the differentiation formulas of the algebraic functions, the differentiation formulas of the remaining functions, that is, tan u, cot u, sec u and csc u may be obtained.
dxduusinucos
dxd
dxduucosusin
dxd
xfu where u cos of Derivative
xfu where u sin of Derivative
xcosxsin
dxd xtan
dxd
xfu where u tan of Derivative
2cosx
xcosdxdsinxxsin
dxdcosx
xtandxd
quotient, of derivative gsinU
xcos
xsinsinxcosxcosx 2
xcos1
xcosxsinxcos 22
22
x secxtandxd 2
dxduusec utandx
d Therefore 2
xtan1
dxd xcot
dxd
xfu where u cot of Derivative
2
2
2 tanxxsec10
tanx
xtandxd10
xtandxd
quotient, of derivative gsinU
xcsc xsin
1
xcosxsinxcos
1
xtanxsec 2
2
2
2
2
2
2
xcsc- xcotdxd 2
dxduucsc- ucotdx
d Therefore 2
xcos1
dxd xsec
dxd
xfu where u sec of Derivative
22 cosx
xsin10 cosx
xcosdxd10
xtandxd
quotient, of derivative gsinU
x secx tan xcos
1xcosxsin
xcosxsin 2
x secx tan xsecdxd
dxduusecutan usecdx
d Therefore
xsin1
dxd xcsc
dxd
xfu where u csc of Derivative
22 x sin
xcos10 x sin
xsindxd10
xcscdxd
quotient, of derivative gsinU
x csc x cot xsin
1xsinxcos
xsinxcos 2
x csc x cot xcscdxd
dxduucscucot- ucscdx
d Therefore
dxdu u cosu sin
dxd
dxdu u sinu cos
dxd
dxdu usecu tan
dxd 2
dxdu ucscu cot
dxd 2
dxdu u sec u tanu sec
dxd
dxdu u csc u cotu csc
dxd
If u is a differentiable function of x, then the following are differentiation formulas of the trigonometric functions
SUMMARY:
A. Find the derivative of each of the following functions and simplify the result:
x3sin2xf .1
xsinexg .2
22 x31cosxh .3
x3cos6
3x3cos2x'f
xsindxdex'g xsin
22x31cosxh
x21xcose xsin
x2
xcose xxx
x2xcosex'g
xsinxsin
x6x31sinx31cos2x'h 22
22 x31sinx31cos2x6
2sinxcosx2xsinfrom
2x312sinx6x'h
3x4cos3x4sin3y .4
233233 x12x4cosx4cosx12x4sinx4sin3'y
xsinxcos2xcos from 22
32 x42cosx36'y
32 x8cosx36'y
x2xtan2xf .5
121
2xsec2x'f 2
12xsecx'f 2
2xtanx'f 2
x1
xtan3logxh .6
22
3 x11x1x1
x1xsecelog
x1xtan
1x'h
x1xcos
1
x1xsin
x1xcos
x1xx1elogx'h
22
3
2
2
x1xcos
x1xsin
1x1
elogx'h 2
3
x1xcos
x1xsin2
1x1
elog2x'h 2
3
x1x2sin
1x1
elog2x'h 2
3
x1x2csc
x1elog2
x'h 23
x2cosx2secy .7
x2seclnx2cosyln
x2seclnyln
sidesboth on ln Applyx2cos
2x2sinx2secln2x2tanx2secx2sec
1x2cos'yy1
ationdifferenti clogarithmi By
x2seclnx2sin2x2cosx2sin2x2cos'y
y1
x2secln1x2sin2
yx2secln1x2sin2'y
x2cosx2secx2secln1x2sin2'y
xcot1
xcot2xh .8 2
22
222
xcot11xcscxcot2xcot21xcsc2xcot1x'h
xcot1xcot2xcot1
xcsc2x'h 2222
2
1xcotxcsc
xcsc2 222
2
1
xsinxcosxsin2
xcsc1xcot2x'h 2
22
2
2
xsinxsinxcosxsin2x'h 2
222
x2cos2x'h
1xcscxF .9 3
1xcsc2
x31xcot1xcscx'F3
233
1xcsc2
1xcsc1xcot1xcscx3x'F 3
3332
1xcsc1xcotx23x'F 332
Find the derivative and simplify the result.
3x45sinlnxh .1
3 2xlncosxf .2
x4cos2
x4sinxg .3
x2cosx4sin2x2sinxcos2xF .4
xcos31siny .5
3
x tanxsinxF .6
yxtany .7
2
2
x1x2cotxF .8
0xyxycot .9
EXERCISES:
0ycscxsec .10 22