lesson 11 derivative of trigonometric functions
TRANSCRIPT
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DIFFERENTIATION OF TRIGONOMETRIC
FUNCTIONS
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OBJECTIVES:• differentiate functions involving trigonometric
functions;• solve problems involving differentiation of
trigonometric functions; and• apply trigonometric identities to simplify the
functions.
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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.
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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
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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
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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
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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
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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
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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
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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:
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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
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3x4cos3x4sin3y .4
233233 x12x4cosx4cosx12x4sinx4sin3'y
xsinxcos2xcos from 22
32 x42cosx36'y
32 x8cosx36'y
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x2xtan2xf .5
121
2xsec2x'f 2
12xsecx'f 2
2xtanx'f 2
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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
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x2cosx2secy .7
x2seclnx2cosyln
x2seclnyln
sidesboth on ln Applyx2cos
2x2sinx2secln2x2tanx2secx2sec
1x2cos'yy1
ationdifferenti clogarithmi By
x2seclnx2sin2x2cosx2sin2x2cos'y
y1
x2secln1x2sin2
yx2secln1x2sin2'y
x2cosx2secx2secln1x2sin2'y
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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
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1xcscxF .9 3
1xcsc2
x31xcot1xcscx'F3
233
1xcsc2
1xcsc1xcot1xcscx3x'F 3
3332
1xcsc1xcotx23x'F 332
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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