CAB-Chp-6-2-NTs - (c) y = ln|53x 2 | Derivative of lnu is...

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CalcAB Chp 6-2 U-Substitution Anti-derivative process for undoing derivatives that used chain rule deriviative
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1. REM: Chain rule derivative 1. Ex(use chain rule derivative) Find each derivative. (a) y = (3x 3 + 1) 5 (b) y = e 4x+3 (c) y = ln|5−3x 2 | chain rule derivative 2 methods (1) use w and f(w) or (2) use inside and outside only
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1. Ex(use chain rule derivative) Find each derivative. (a) y = (3x 3 + 1) 5 chain rule derivative 2 methods (1) use w and f(w) or (2) use inside and outside only Method 1 inside w = 3x 3 + 1 f(w) = w 5 w’ = 9x 2 f’(w) = 5w 4 chain rule = 9x 2 5w 4 f’(x) = 9x 2 5(3x 3 + 1) 4 f’(x) = 45x 2 (3x 3 + 1) 4 Method 2 deriv. of inside deriv. of outside f’(x) = 9x 2 5(3x 3 + 1) 4 f’(x) = 45x 2 (3x 3 + 1) 4
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1. Ex(use chain rule derivative) Find each derivative. (b) y = e 4x+3 Method 1 inside w = 4x+3 f(w) = e w w’ = 4 f’(w) = e w chain rule = 4 e w f ’( x) = 4 e 4x+3 f ’( x) = 4 e 4x+3 Method 2 deriv. of inside deriv. of outside f’(x) = 4 e 4x+3 f’(x) = 4 e 4x+3
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1. Ex(use lnu derivative) Find each derivative.
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Unformatted text preview: (c) y = ln|53x 2 | Derivative of lnu is u'/u where u is the value value u = 53x 2 u = 6x f(x) = The process: Change the integral given in terms of x ( ) to an equivalent, simpler integral in terms of u ( ) 2. Integrals requiring u-substitution The plan is to replace the part of an integral that makes it impossible to get the integral. The complicated part is replaced with u and a new, simpler integral is made. Then get the anti-derivative F(u) this is in terms of u Convert F(u) back to terms of x so the integral is F(x) + C 3. u can substitute for different parts Integrate each. (d) (e) (f) Ex( Let u be inside ( ) s ) Integrate each. (g) (h) What is wrong with this problem such that it cannot be done using u-substitution? Ex( Let u be exponent of e u ) Integrate each. (i) (j) Ex( Let u be the dominator of 1/u ) Integrate each. (k) (j) Ex( Let u be lnx )...
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CAB-Chp-6-2-NTs - (c) y = ln|53x 2 | Derivative of lnu is...

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