MAT 21A Lecture Notes - Lecture 10: Product Rule, Power Rule, Quotient Rule
MAT 21A – Lecture 10 – Differentiation Rules Continued
• Problem: Find
Lets attept this prole i two differet was.
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Or
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12x11 5x4 * 7x6 = 35x10
12x11 35x10
• Another method to differentiate functions where multiplication is involved is
called the product rule, which states:
where we can call one of the factors f(x) and the other g(x). In the
above example, is f(x) and is g(x). With the product rule applied we have,
= (5x4)(x7) + (x5)(7x6) = 5x11 + 7x11 = 12x11. This is the correct answer.
• Even though in this problem, it was easy to rewrite as equivalent to ,
there are other cases where you cannot do that so the product rule is necessary.
• Example 1: Evaluate:
Method #1: Multiply everything out first and then differentiate. Although that
may not be too difficult, it would still be time consuming.
Method #2: Apply the product rule, which gives us:
(6x + 8)(9x3 – 8x2 + 17x +9) + (3x2 + 8x + 5)(27x2 – 16x + 17)
• A function, f(x) such as x2sinx cannot be simplified or rewritten. In other words,
no algebraic manipulation or multiplying factors can be done.
• Proof of Product Rule
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= fxg’x + gxf’x
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MAT 21A Full Course Notes
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Document Summary
Let(cid:859)s atte(cid:373)pt this pro(cid:271)le(cid:373) i(cid:374) two differe(cid:374)t wa(cid:455)s. Mat 21a lecture 10 differentiation rules continued: another method to differentiate functions where multiplication is involved is, problem: find (cid:3031)(cid:3031)(cid:4666)(cid:2873) 7(cid:4667) (cid:3031)(cid:3031)(cid:4666)(cid:2873) 7(cid:4667) = (cid:3031)(cid:3031)(cid:4666)(cid:2869)(cid:2870)(cid:4667) (cid:3031)(cid:3031)(cid:4666)(cid:2873) 7(cid:4667) = (cid:3031)(cid:3031)(cid:4666)(cid:2873)(cid:4667) * (cid:3031)(cid:3031)(cid:4666)7(cid:4667) 12x11 35x10 called the product rule, which states: (cid:3031)(cid:3031)[(cid:1858)(cid:4666)(cid:4667)(cid:1859)(cid:4666)(cid:4667)]= (cid:1858) (cid:4666)(cid:4667)(cid:1859)(cid:4666)(cid:4667)+ (cid:1858)(cid:4666)(cid:4667)(cid:1859) (cid:4666)(cid:4667) where we can call one of the factors f(x) and the other g(x). In the above example, (cid:2873) is f(x) and 7 is g(x). With the product rule applied we have, (cid:3031)(cid:3031)(cid:4666)(cid:2873) 7(cid:4667) = (5x4)(x7) + (x5)(7x6) = 5x11 + 7x11 = 12x11. This is the correct answer: even though in this problem, it was easy to rewrite (cid:2873) 7 as equivalent to (cid:2869)(cid:2870), example 1: evaluate: (cid:3031)(cid:3031)(cid:4666)(cid:885)(cid:2870)+(cid:890)+5(cid:4667)(cid:4666)(cid:891)(cid:2871) (cid:890)(cid:2870)+(cid:883)(cid:889)+(cid:891)(cid:4667) Method #1: multiply everything out first and then differentiate. Although that there are other cases where you cannot do that so the product rule is necessary. may not be too difficult, it would still be time consuming.
