MATH114 Lecture Notes - Lecture 17: Quotient Rule

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ultramarineeel54

13 Oct 2018

School

University of Alberta

Department

Mathematics

Course

MATH114

Professor

Powell,Elizabeth

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MATH114 Lecture Notes - Lecture 16: Power Rule

MATH114 Lecture Notes - Lecture 17: Quotient Rule

MATH114 Lecture 18: MATH114-LECTURE#18-CHAIN RULE

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For example, find the derivative of (cid:2871)47 (cid:1856)(cid:1856)(cid:1876)[(cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1859)(cid:4666)(cid:1876)(cid:4667)]= (cid:1859)(cid:4666)(cid:1876)(cid:4667)(cid:1856)(cid:1856)(cid:1876)(cid:1858)(cid:4666)(cid:1876)(cid:4667) (cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1856)(cid:1856)(cid:1876)(cid:1859)(cid:4666)(cid:1876)(cid:4667) (cid:1859)(cid:2870)(cid:4666)(cid:1876)(cid:4667) (cid:4678)(cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1859)(cid:4666)(cid:1876)(cid:4667)(cid:4679) = (cid:1858) (cid:4666)(cid:1876)(cid:4667)(cid:1859)(cid:4666)(cid:1876)(cid:4667) (cid:1858)(cid:4666)(cid:1876)(cid:4667)(cid:1859) (cid:4666)(cid:1876)(cid:4667) (cid:1859)(cid:2870)(cid:4666)(cid:1876)(cid:4667) (cid:1864)(cid:1857)(cid:1872) (cid:1858)(cid:4666)(cid:1876)(cid:4667)=(cid:885)(cid:1876)(cid:2872),(cid:1872) (cid:1873)(cid:1871) (cid:1858) (cid:4666)(cid:1876)(cid:4667)=(cid:883)(cid:884)(cid:1876)(cid:2871) (cid:1864)(cid:1857)(cid:1872) (cid:1859)(cid:4666)(cid:1876)(cid:4667)=7,(cid:1872) (cid:1873)(cid:1871) (cid:1859) (cid:4666)(cid:1876)(cid:4667)=7ln (cid:4666)7(cid:4667) (cid:1859)(cid:2870)(cid:4666)(cid:1876)(cid:4667) (cid:4666)7(cid:4667)(cid:2870) lim 0sin(cid:4666)(cid:1876)(cid:4667)(cid:1876) =(cid:883) lim 0(cid:4666)cos(cid:4666)(cid:1876)(cid:4667) (cid:883)(cid:4667) (cid:1876) You can use the two limits and trig identities to find the derivatives of trig functions by definition. E. g. find the derivative of sin(x) using its definition lim 0(cid:1858)(cid:4666)(cid:1876)+ (cid:4667) (cid:1858)(cid:4666)(cid:1876)(cid:4667) lim 0sin (cid:4666)(cid:1876)+ (cid:4667) sin (cid:4666)(cid:1876)(cid:4667) lim 0sin(cid:4666)(cid:1876)(cid:4667)cos(cid:4666) (cid:4667)+sin(cid:4666) (cid:4667)cos (cid:4666)(cid:1876)(cid:4667) sin (cid:4666)(cid:1876)(cid:4667) lim 0sin(cid:4666)(cid:1876)(cid:4667)cos(cid:4666) (cid:4667) sin (cid:4666)(cid:1876)(cid:4667)+sin(cid:4666) (cid:4667)cos (cid:4666)(cid:1876)(cid:4667) lim 0sin(cid:4666)(cid:1876)(cid:4667)(cid:4666)cos(cid:4666) (cid:4667) (cid:883)(cid:4667)+sin(cid:4666) (cid:4667)cos(cid:4666)(cid:1876)(cid:4667) Note: an easy way to remember them is to memorize the left side and know that the right side is just the negative opposite (opposite of sine is cosine). You can essentially use the sine and cosine derivatives to find the derivatives of the other functions. For example, find the derivative of tan(x): f(cid:4666)x(cid:4667)= tan(cid:4666)(cid:1876)(cid:4667)=sin(cid:4666)x(cid:4667) cos(cid:4666)(cid:1876)(cid:4667) (cid:1858) (cid:4666)(cid:1876)(cid:4667)=(cid:1856)(cid:1856)(cid:1876)sin(cid:4666)(cid:1876)(cid:4667) cos(cid:4666)(cid:1876)(cid:4667) sin(cid:4666)(cid:1876)(cid:4667) (cid:1856)(cid:1856)(cid:1876)cos (cid:4666)(cid:1876)(cid:4667) (cid:4666)cos(cid:4666)(cid:1876)(cid:4667)(cid:4667)(cid:2870) (cid:1858) (cid:4666)(cid:1876)(cid:4667)=cos(cid:4666)x(cid:4667)cos(cid:4666)x(cid:4667)+sin(cid:4666)x(cid:4667)sin (cid:4666)x(cid:4667)

Related Questions

3. (21 pts) Evaluate the limit or say that the limit does not exist. If a limit does not exist, explain why. Show your work. You may NOT use a table of values, a graph, or L'Hospitals's Rule to justify your answer. (Note: Possible answers include + or -..) (a) (7 pts) x2 - 2x - 3 - lim x+3 Vx+1 - 2 x2 - 2x - 3 Vx+ 1 + 2 lim x+3 Vx+1 -2 Vx+1 +2 (x - 3) (x + 1) (Vx+ = lim x+3 x + 1 - 4 = lim x+3 (x - 3) (x + 1) (Vx+1 +2). -- x-3 = lin (x + 1) (Vx+1 +2) = (3 + 2) (V3+1 +2) = = 4 (V4 +2) = 16 (b) (7 pts) lim xsin (In x) X >0+ NOTE: We cannot apply the Product Law, since lim sin (1n x) DOES NOT EXIST. On the other hand, -1s sin (in x) s 1 and , therefore, -xs x sin (in x) sx, (since x > 0). Since lim X = lim (-X) = 0, x0+ X0 the inequality above implies that lim xsin (in x) = 0, by the Squeeze Theorem . x0+ EXPLANATION : Since lim ln x = -0, x →0+ sin (lnx) oscillates between 1 and -1 as x + 0*, and, therefore lim sin (In x) DOES NOT EXIST. x >0+ (c) (7 pts) in x+2- (lnx) +In 20 -= -00 (x - 2) 70-