SS_Mat135_T1

# 3 2 solution exponent laws b m b n b m n b b

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Unformatted text preview: erivative of both the numerator and the denominator. 3 2 lim x + 3 x − 10 x 3 x − 3x + 2 x d3 x + 3 x 2 − 10 x dx = lim = x→2 d x 3 − 3x 2 + 2 x dx x→2 ( ) ( () ) 2 −5+ f (x + h ) − f (x ) . h EXAMPLE: Find the derivative of the function 2 f (x ) = (x + 1) by first principles. SOLUTION: f (x + h ) − f (x ) h (x + h + 1)2 − (x + 1)2 = lim h →0 h f ′( x ) = lim = lim (x 2 h→0 )( ) + 2 xh + 2 x + h 2 + 2h + 1 − x 2 + 2 x + 1 h 2 xh + h 2 + 2h h(2 x + h + 2 ) = lim h →0 h→0 h h = lim 2 x + h + 2 = 2(x + 1) = lim h →0 TECHNIQUES FOR FINDING LIMITS: First try substitution, factoring If you’re taking the limit as x → ∞ of a quotient, and substitution yields an invalid answer, then try first to divide both the numerator and the denominator of the limit expression by its highest power of x If the limit expression contains absolute values, then try breaking the limit up into two one-sided limits If the limit expression is a quotient, and if factoring doesn’t work, then try l’Hopital’s Rule RULES FOR DIFFERENTIATION: For functions u,v and for constants c, n: () dy du , cu n = cnu n −1 dx dx dy dv du , (uv ) = u + v dx dx dx du dv v −u dy ⎛ u ⎞ dx dx ⎜ ⎟= dx ⎝ v ⎠ v2 DERIVATIVES TO REMEMBER: EXAMPLE: x2 − 5 − 1+ x lim there exists a δ &gt; 0 such that |f(x) - L| &lt; ε for all...
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## This note was uploaded on 03/15/2010 for the course MAT MAT135 taught by Professor Treung during the Fall '08 term at University of Toronto- Toronto.

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