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Bob Miller's High School Calc for the Clueless - Honors and - download pdf or read online

Posted On April 20, 2018 at 2:09 pm by / Comments Off on Bob Miller's High School Calc for the Clueless - Honors and - download pdf or read online

By By (author) Bob Miller

ISBN-10: 0071488456

ISBN-13: 9780071488457

ISBN-10: 0071594639

ISBN-13: 9780071594639

With Bob Miller at your facet, you by no means must be clueless approximately math back!

Algebra and calculus are tricky on highschool scholars such as you. Professor Bob Miller, with greater than 30 years' educating event, is a grasp at making the advanced basic, and his now-classic sequence of Clueless learn aids has helped tens of millions comprehend the harsh subjects.

Calculus-with its integrals and derivatives-is recognized for tripping up even the fastest minds. Now Bob Miller-with his 30-plus years' event instructing it-presents highschool calculus in a transparent, funny, and interesting way.

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Read or Download Bob Miller's High School Calc for the Clueless - Honors and AP Calculus AB & BC PDF

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Extra resources for Bob Miller's High School Calc for the Clueless - Honors and AP Calculus AB & BC

Sample text

If we had a complicated function, it would take forever. We will list the rules, interpret them, and give examples. Proofs are found in most calculus books. RULE 1 If f(x) ϭ c, then fЈ(x) ϭ 0. The derivative of a constant is 0. RULE 2 If f(x) ϭ x, then fЈ(x) ϭ 1. RULE 3 If f(x) ϭ xn, then fЈ(x) ϭ nxnϪ1. Bring down the exponent and subtract 1 to get the new exponent. EXAMPLE 6— A. If y ϭ x6, then yЈ ϭ 6x5. B. If y ϭ xϪ7, then yЈ ϭ Ϫ7xϪ8. 5 then yЈ ϭ x2/3. 3 then yЈ ϭ ␲x␲Ϫ1. C. If y ϭ x5/3, D. If y ϭ x␲, Let us prove the power rule for positive integers.

If h ϭ 63°, the area of the sector ϭ 63/360 ϫ ␲r2. In general the area would be h/360 ϫ ␲r2. Angle h is in radians. The area is h/2␲ ϫ ␲r2. Since r ϭ 1, the area of this sector is h/2. The Basics Now the area of ⌬OAC р sector OAB р ⌬OAD or 1 1 1 Q R(cos h)(sin h) р Q R h р Q R AD 2 2 2 To get AD, we note that the smaller and larger triangles are similar; we get y sin h AD AD x ϭ cos h ϭ OA ϭ 1 ϭ AD sin h , multiply the whole inequality cos h by 2, and divide by sin h. Replace AD by We get cos h р 1 h р sin h cos h Taking limits, we get lim cosh р lim hS0 hS0 1 h 1 h р lim or 1 р lim р1 h S 0 sin h sinh h S 0 cos h We also notice that the only way that u р v and at the h ϭ 1.

Some you know; others you need to review; and still others are new. All are important. Let’s show that if f(x) ϭ sin x, then fЈ(x) ϭ cos x. As usual, we must use the definition of derivative. lim hS0 sin(x ϩ h) Ϫ sin x f(x ϩ h) Ϫ f(x) ϭ lim hS0 h h sin (A ϩ B) ϭ sin A cos B ϩ cos A sin B ϭ lim hS0 sin x cos h ϩ cos x sin h Ϫ sin x h Factoring out sin x from the first and third terms, and splitting by using short division, we get lim hS0 sin x (cos h Ϫ 1) sin h ϩ lim cos x h S 0 h h The Basics If we could show lim hS0 cos h Ϫ 1 sin h ϭ 0 and lim ϭ 1, hS0 h h our theorem would be proved.

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Bob Miller's High School Calc for the Clueless - Honors and AP Calculus AB & BC by By (author) Bob Miller


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