1 + t5 dt with respect to x. du = 2. dg du = f(u). du dx. dg dx = dg. du du. dg du. dx = 4x3. - page 1 -

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Claude Riley
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1 Eercise. Find te derivative of g( 3 + t5 dt wit respect to. Solution: Te integrand is f(t + t 5. By FTC, f( + 5. Eercise. Find te derivative of e t2 dt wit respect to. Solution: Te integrand is f(t e t2. By FTC, 2 e. Let s just assume tat f(t is continuous every- Fundamental Teorem of Calculus, rewritten wit u were (just to ave fewer tecnicalities. Fi a number called a. Define a function g wit input u by g respect to u is given by Eercise. Find te derivative of k f(u. Te integrand is f(t e t2. By FTC, dk 2 e u. a f(t dt. Ten, te derivative of g wit e t2 dt wit respect to u using te rewritten FTC. Cain Rule Formula, rewritten. Eercise. Find te derivative of g( 4 sec t dt wit respect to. Solution: Let u 4. Ten, g can be rewritten as g te Cain Rule, so we need to find Finding and and., sec t dt. We were asked to find. By is just like te set up of te previous eercise. By using te FTC rewritten wit te u, is more routine: Putting tis all togeter: sec u. 43. Cain ( sec u (4 3 sec( {{{{ - page -

2 Eercise. Find te derivative of g( sin 9 e t2 dt wit respect to. Solution: Let u sin. Ten g can be rewritten as g ( e u2 {{ ( cos 9 e t2 dt. Like te last problem, e (sin 2 cos. wic can also optionally be rewritten as e sin2 cos or cos e sin2. Eercise. Find te derivative of g( log 3 + cos t dt wit respect to. Solution: Te function is te lower endpoint of integration, so rewrite g as follows: g log3 + cos t dt. Tings can start to get confusing wit te minus sign, so I d like to set up a new variable to represent all of wat s in g ecept te minus sign and I ll call it. To be a bit clearer: log3 g + cos t dt. In oter words, so tat g. And so. Using u log 3, rewrite like tis: log3 + cos t dt. Tus, + cos t dt. Cain ( ( + cos(log3 + cos u. {{ {{ ln 3 ln 3 To finis, recall we already said tat. Tus, + cos(log3. ln 3 - page 2 -

3 Eercise. Find te derivative of g( log 3 + cos t dt wit respect to. Solution: Te integral as bot endpoints of integration being functions, so we need to rewrite g by splitting te integral into two separate integrals: g log 3 log 3 + cos t dt + cos t dt + + cos t dt + cos t dt + + cos t dt + cos t dt log 3 log3 + cos t dt. k Since g k, te using te Difference Rule for differentiation, dk. Let u and v log 3. (Note tat I m using two DIFFERENT letters: u and v, just for good bookkeeping. Using tese coices for u and v, we can rewrite bot and k. Tey are rewritten like tis: v + cos t dt and k + cos t dt. Ten, dk {{ dk {{ dk ( + cos u ( 5 ( ( + cos v {{ dk All tat s left to do is replace u and v, since we introced tem: ( ( + cos( 5 ( + cos(log3 ( ln 3 ln 3 {{ - page 3 -

4 Eercise. Find te derivative of g( wit respect to. Solution: First rewrite g. Since tere s an in bot endpoints of integration, we ll again ave to rewrite te integral as te sum of two integrals. g + m Rewrite by using u 5. Ten we get: Note tat because m is an integral wic already as an at te top, tere is no need to replace te wit a v. Since g m, we ave dm {{ dm ( 3 + ln u ( 5 ln 5 ( ( 3 + ln 3 + ln(5 5 {{ ln ln dm How many variable substitutions are required? Note again tat tere is NO NEED to rewrite m by saying tat v and ten writing m v It is by no means wrong to do tis, but note tat you ll write dm anyway! Again, it s not WRONG, but tis is a waste of time! dm and tat is just More importantly, if you just do tis substitution witout tinking, ten you ve sort of missed te point of wy we did tese substitutions in te first place!! Don t turn tis into a routine. Te use of u ad a purpose! For eample, go back to te first tree eercises: Find te derivative of g( Find te derivative of Find te derivative of k 3 + t5 dt wit respect to. e t2 dt wit respect to. e t2 dt wit respect to u. You did not do a variable substitution in ANY of tese! (You did not need te Cain Rule for any of tese! In same way, look at m. If te FIRST eercise on tis practice seet was Find te derivative of m you would NOT ave said u or v. wit respect to - page 4 -

5 ( z Eercise. Find te derivative of (z tan + m2 dm wit respect to z. Solution: Let g Cain Rule, z + m2 dm so ten tan(g. Te question asks us to find. By te. Since we need to find, let s do tat first. Note tat were tere is traditionally a t, tere is now an m. Were tere s traditionally an, tere s currently a z. But all te same ideas still apply, just wit different letters! Let u z. We can rewrite g like tis: So g + m2 dm ( + u 2 (z 5 ( + (z (z 2 5 {{{{ And tus, ( ( sec 2 (g ( [ ( z + (z 2 (z 5 sec 2 {{{{ + m2 dm ] [ ( ] + (z 2 (z 5 Ensure tat your final answer looks like tis. We introced te g and u, so te final answer souldn t ave any of tese. Te final answer sould only be in terms of z and m. - page 5 -

Continuity and Differentiability Worksheet

Continuity and Differentiability Workseet (Be sure tat you can also do te grapical eercises from te tet- Tese were not included below! Typical problems are like problems -3, p. 6; -3, p. 7; 33-34, p. 7;