## 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.

**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 xϪ1. 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.

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

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