Math1011_Week09

Math1011_Week09 - Math1011 Section 3.7 Consider the curve...

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Math1011 Section 3.7 10/18/2010 22 Consider the curve x y - 6y + 2 = 0 Find the equation of the tangent line to the curve at point (2,1) 23 Given 5 tan re θ θθ =+ + dr d Find d dr Find

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Math1011 Section 3.7 1 22 Consider the curve x y - 6y + 2 = 0 Find the equation of the tangent line to the curve at point (2,1) 2 2 2 2 dd d d dx dx dx dx 2 26 2(2)(1) 44 86 2 (2) 2(1) 6 Implicit Differentiation (x y ) (6 ) (2) (0) () 6 ' 00 2' 2 6 ' 0 6 ' 2 '( 2 6) 2 ' at point (2, 1), y'= 2 dx xy y yx yyx x yy −− −+ = +− + = = −= = == = 11 12 (2 ) equation of the line tangent to the curve at (2, 1) is 1 2( 2) mx x =− 23 Given 5 tan re θ θθ =+ + dr d Find d dr Find 5t a n 10 sec 3 10 sec 3 d dr e + + + 1 10 sec 3 a n 0 s e c 3 1( 1 0 s e c 3 ) ++ + + + = (Doc #1011w.37.01)
Math1011 Section 3.8 2 -1 2 Find (f )'(3) if f(x) = x 8+x 2 1 d 1 dx 1 Show that (tan ) x + =

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Math1011 Section 3.8 2 -1 2 Find (f )'(3) if f(x) = x 8+x 1 1 1 '( ( )) -1 2 1/2 2 1 2 21 / 2 2 10 11 23 3 -1 3 1 10 10/3 () ' ( ) (3) 1 and '( ) ( )(8 ) (2 ) 8 '(1) (1)( )(8 (1) ) (2)(1) 8 (1) 3 (f )'(3) ff x fx ff x f = == + + + =+ + + = + = -1 Since (1) 3 implies that 1 2 1 d 1 dx 1 Show that (tan ) + = 1 2 22 2 2 1 1 '( ( )) 1 1 sec (tan ( )) 1 1 sec ( ) 1 1 (1 ) / 1 1 1 1 () t a n '( ) sec ' ( ) (tan ) (tan ) (tan ) (tan ) d dx θ + + = = = = = = = θ x²+1 x 1 (Doc #1011w.38.01)(Adams 3.1.30)
Math1011 Section 3.8 3 Differentiate the functions 2 () ln ( s in) Ft t =+ 21 23 Find f'(t) if ( ) ln ft + + =

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Math1011 Section 3.8 3 Differentiate the functions 2 () ln ( s in) Ft t =+ 2 2 2 2 d 1 dt 2 1 sin 1 sin 2c o s sin '( ) ln( sin ) Note: (ln ) '( ) ( sin ) '( ) (2 cos ) '( ) d dt tt + + + + =⋅+ = = 21 23 Find f'(t) if ( ) ln ft + + = 1/2 1 2 1 2 11 1 221 1 1 1 1 () ( ( ) ) ( ) (ln(2 1) ln(2 3)) '( ) (ln(2 1) ln(2 3)) '( ) [ 1) 3)] '( ) [ (2) (2)] '( ) [ ] dd + + + + ++ = = + + =− = + (Doc #1011w.38.02)
Math1011 Section 3.8 4 sin Find the derivative of x y = (x+2) Find the derivative of y=(x+1)

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Math1011 Section 3.8 4 sin Find the derivative of x y = sin 1 1 11 ln ln ln sin ln move exponent down ' (sin ln ) product rule y' [sin (ln ) ln (sin )] '[ s i n( )l d dx dd yx xx = = = =+ 1 sin 1 c o s) ] ' [ s i c o ] ' [ s i c o ] yy (x+2) Find the derivative of y=(x+1) (x+2) d 1 y dx 1 y dx dx y x+1 x+2 x+1 (x+2) x+2 x+1 lny =ln (x+1) y = (x+2)ln (x+1) y' = [(x+2)ln (x+1)] y' = [( 2) ln (x+1)+ln(x+1) ( 2)] y' = [( 2)( )+ln(x+1)] y' = y[ +ln(x+1)] y' = (x+1) [ +ln(x+1)] ++ + (Doc #1011w.38.03)
Math1011 Section 3.11 5 3 Find the linearization of ( ) at -8 and use this to find the value of L(-7.9) fx x a = = Find the linearization of ( ) sin( ) at 0 and use this to find the value of L(0.1) = =

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Math1011 Section 3.11 5 3 3
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This note was uploaded on 12/06/2010 for the course MATH 1110 taught by Professor Martin,c. during the Fall '06 term at Cornell University (Engineering School).

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Math1011_Week09 - Math1011 Section 3.7 Consider the curve...

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