November 12, 1997
-963295-15938500 Sec 2.5 – General Differentiation Differentiation of Sine & CosineName:(1) Let’s try to prove the derivative of the function fx=sinx using the definition of the derivative.f'x=limh→0 sinx+h-sinxhWe could use the sum and difference trig identities to substitute sinx+h=sinxcosh+cosx sinh f'x=limh→0 sinxcosh+cosx sinh-sinxhWe can rearrange the terms of the numeratorf'x=limh→0 sinxcosh-sinx+cosx sinhhWe could then factor out sinx from the first two terms.f'x=limh→0 sinxcosh-1+cosx sinhhNext, we could use limit laws to rewrite the following statement:f'x=sinx?limh→0 cosh-1h + cosx?limh→0 sinhh(A) Which leaves us with two indeterminate form limits. We will need to use the Squeeze Limit Theorem. Let’s first investigate the limit limh→0sinhh. Consider using the following diagram for which 0≤h<π2 .373824538100036576001054100-81915073025Arc Length BD ≤ BC h ≤ tanh h ≤ sinhcosh h?cosh ≤ sinh cosh ≤ sinhhArc Length=r?θLength BD=1?hLength BD=h tanθ=OPPADJtanh=BC1tanh=BC DE ≤ Arc Length BD sinh ≤ h sinhh ≤ hh sinhh ≤ 1Arc Length=r?θLength BD=1?hLength BD=h sinθ=OPPHYPsinh=DE1sinh=DEArc Length BD ≤ BC h ≤ tanh h ≤ sinhcosh h?cosh ≤ sinh cosh ≤ sinhhArc Length=r?θLength BD=1?hLength BD=h tanθ=OPPADJtanh=BC1tanh=BC DE ≤ Arc Length BD sinh ≤ h sinhh ≤ hh sinhh ≤ 1Arc Length=r?θLength BD=1?hLength BD=h sinθ=OPPHYPsinh=DE1sinh=DE138430073025128270016954500256540013144500-1079501263650036385502540Although we were only working with a right-handed limit we could find the left hand limit by substituting h with – h. lim h→0-sinhh=lim h→0+sin-h-h Since the function fx=sinx is an odd functionsin-x=-sinxIf we make the appropriate substitution: lim h→0+sin-h-h=lim h→0+-sinh-h=lim h→0+sinhh=1 00Although we were only working with a right-handed limit we could find the left hand limit by substituting h with – h. lim h→0-sinhh=lim h→0+sin-h-h Since the function fx=sinx is an odd functionsin-x=-sinxIf we make the appropriate substitution: lim h→0+sin-h-h=lim h→0+-sinh-h=lim h→0+sinhh=1 -615950231140 cosh ≤ sinhh ≤ 1 limh→0+ cosh ≤ lim h→0+sinhh ≤ limh→0+ 1 1 ≤ lim h→0+sinhh ≤ 1By the Squeeze Limit Theorem:lim h→0+sinhh=100 cosh ≤ sinhh ≤ 1 limh→0+ cosh ≤ lim h→0+sinhh ≤ limh→0+ 1 1 ≤ lim h→0+sinhh ≤ 1By the Squeeze Limit Theorem:lim h→0+sinhh=1center380010M. Winking ? Unit 2-5 page 3700M. Winking ? Unit 2-5 page 37(B) Next, let’s investigate the limit limh→0cosh-1h and consider multiplying it by something equivalent to 1. limh→0 cosh-1h ?cosh+1cosh+1 After expanding the numerator, we would have.=limh→0 cos2h-1h?cosh+1 Then, we could use the trigonometric Pythagorean identity sin2h+cos2h=1 rearranged to cos2h-1=-sin2h = limh→0 -sin2hh?cosh+1 Next, we could separate the fraction into 2 pieces using limit laws: = limh→0 sinhh?limh→0 -sinhcosh+1 The limit on the left we just determined was limh→0sinhh=1 and the limit on the right we could use direct substitution: limh→0cosh-1h = 1?-sin0cos0+1= 1?01+1=0 Finally, we can go back to the original limit with the two indeterminate limits limh→0cosh-1h=0 and limh→0sinhh=1: f'x=sinx?limh→0 cosh-1h + cosx?limh→0 sinhhf'x= sinx ? 0 + cosx ? 1 f'x=cosx -749300128905(2) Let’s try to prove the derivative of the function gx=cosx using the definition of the derivative. g'x=limh→0 cosx+h-cosxh We could use the sum and difference trig identities to substitute cosx+h=cosxcosh-sinx sinh g'x=limh→0 cosxcosh-sinx sinh-cosxhWe can rearrange the terms of the numeratorg'x=limh→0 cosxcosh-cosx-sinx sinhhWe could then factor out cosx from the first two terms. g'x=limh→0 cosxcosh-1-sinx sinhhNext, we could use limit laws to rewrite the following statement: g'x=cosx?limh→0 cosh-1h- sinx?limh→0 sinhh Finally, we can go back to the original limit with the two indeterminate limits limh→0cosh-1h=0 and limh→0sinhh=1:center675607M. Winking ? Unit 2-5 page 3800M. Winking ? Unit 2-5 page 38 g'x=cosx?limh→0 cosh-1h- sinx?limh→0 sinhh=cosx?0- sinx?1=- sinx1733138-97790Exponential RuleGiven: fx=ax,where a is a constant.f'x=lna?ax0Exponential RuleGiven: fx=ax,where a is a constant.f'x=lna?ax4171950-95250Natural ExponentGiven: fx=ex,where e ≈2.7182818.f'x=ex00Natural ExponentGiven: fx=ex,where e ≈2.7182818.f'x=ex-721772-101600Power RuleGiven: fx=a?xn,where a and n are constants.f'x=n?a?xn-10Power RuleGiven: fx=a?xn,where a and n are constants.f'x=n?a?xn-12641600112395Quotient Rule,where f and g are differentiable.fg'=gxf'x-fxg'xgx200Quotient Rule,where f and g are differentiable.fg'=gxf'x-fxg'xgx2-196850110902Product Rule,where f and g are differentiable.f?g'=fx?g'x+gx?f'x0Product Rule,where f and g are differentiable.f?g'=fx?g'x+gx?f'x3060700168052 Derivative of CosineGiven: fx=cosxf'x=-sin x Derivative of CosineGiven: fx=cosxf'x=-sin x-1089313162560 Derivative of SineGiven: fx=sinxf'x=cos x Derivative of SineGiven: fx=sinxf'x=cos xleft218787Using the various methods shown determine the general derivative of the following: fx=3sinx+x2B. fx=ex?cosxC. y=x2?cosxD. y=cosxx2E. fx=cosx?sinxF. y=3xcosxcenter504702M. Winking ? Unit 2-5 page 3900M. Winking ? Unit 2-5 page 39Given fx=x2?cosx, determine the exact value of f'π6.Given fx=xsinx , determine the exact value of f'π2.Given fx=gx?sinx , gπ3=3 , and g'π3=-2, determine the exact value of f'π3.Given fx=cosxgx , gπ4=3 , and g'π4=-2, determine the exact value of f'π4.39257851237800Given fx=qx?sinx , determine the exact value of f'0.center399201M. Winking ? Unit 2-5 page 4000M. Winking ? Unit 2-5 page 40Determine the higher order derivatives b. Find Find . Find the equation of the line tangent to fx=x+sinx at x=4π3.Find the equation of the line tangent to fx=gx?cosx at x=π4 given:gπ4=4g'π4=2center1964690M. Winking ? Unit 2-5 page 4100M. Winking ? Unit 2-5 page 41 ................
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