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### HW 12 - Implicit and Log Differentiation-solutions (1)

Course: M 408, Spring 2012
School: University of Texas
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Word Count: 1118

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(cw27622) whang HW 12 - Implicit and Log Differentiation bormashenko (55335) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Thus 3x2 + 6x dy = , dx 2y 1 and so at the point (1, 2) the tangent line has 3 (1)2 + 6 (1) 9 slope = = . 2 (2) 4 Now we can use the point-slope formula to find an equation for...

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(cw27622) whang HW 12 - Implicit and Log Differentiation bormashenko (55335) This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Thus 3x2 + 6x dy = , dx 2y 1 and so at the point (1, 2) the tangent line has 3 (1)2 + 6 (1) 9 slope = = . 2 (2) 4 Now we can use the point-slope formula to find an equation for the tangent line: y-2 = 9 (x - 1) . 4 The curve with equation y 2 = x3 + 3x2 is called a Tschirnhausen cubic. Its graph looks like y After simplification this becomes y = x 002 Find dy when dx tan(x - y) = x - 3y . Find an equation for the tangent line to this curve at the point (1, 2). 1. y = 2. y = 3. y = 3 17 x- 4 4 1 9 x- 2 4 1 9 x - correct 4 4 3 - sec2 (x - y) dy = dx sec2 (x - y) - 1 dy 3 - sec2 (x - y) = dx sec2 (x - y) + 1 3 + sec2 (x - y) dy = dx sec2 (x - y) - 1 sec2 (x - y) + 1 dy = dx sec2 (x - y) - 3 sec2 (x - y) - 1 dy = dx sec2 (x - y) + 3 dy sec2 (x - y) - 1 = correct dx sec2 (x - y) - 3 1 9 x- . 4 4 10.0 points 1. 2. 3. 4. 5. 6. 17 3 4. y = - x - 4 4 9 1 5. y = - x + 4 4 Explanation: Differentiating y 2 = x3 + 3x2 implicitly we see that 2y dy = 3x2 + 3 (2x) . dx Explanation: Differentiating implicitly with respect to x, we see that sec2 (x - y) 1 - dy dy = 1-3 . dx dx whang (cw27622) HW 12 - Implicit and Log Differentiation bormashenko (55335) After rearranging, this becomes dy sec2 (x - y) - 3 = sec2 (x - y) - 1 . dx Consequently, dy sec2 (x - y) - 1 . = dx sec2 (x - y) - 3 so dy y = . dx 2y - x - 5 Thus 003 10.0 points dy dx P 2 To determine the tangent lines we need also the value of the derivative at P and Q. But by implicit differentiation, 2y dy dy - (x + 5) - y = 0. dx dx keywords: = 5 , 3 dy dx Q 2 = - . 3 The points P and Q on the graph of y 2 - xy - 5y + 10 = 0 have the same x-coordinate x = 2. Find the point of intersection of the tangent lines to the graph at P and Q. 1. intersect at = 2. intersect at = 3. intersect at = 4. intersect at = 5. intersect at = 20 5 , 7 3 20 5 ,- 7 7 5 20 , 3 7 10 20 , 3 7 5 20 , 7 7 correct By the point-slope formula, therefore, the equation of the tangent line at P is y-5 = while that at Q is 2 y - 2 = - (x - 2). 3 Consequently, the tangent lines at P and Q are y = and 2 10 y+ x = 3 3 respectively. These two tangent lines intersect at = 5 20 , 7 7 . 5 5 x+ 3 3 5 (x - 2), 3 Explanation: The respective y-coordinates at P, Q are the solutions of () y 2 - xy - 5y + 10 = 0 004 10.0 points Differentiate the function f (x) = cos(ln 2x) . 1. f (x) = - 2. f (x) = sin(ln 2 x) correct x at x = 2; i.e., the solutions of y 2 - 7y + 10 = (y - 5)(y - 2) = 0. Thus P = (2, 5), Q = (2, 2). 1 cos(ln 2 x) whang (cw27622) HW 12 - Implicit and Log Differentiation bormashenko (55335) 2 sin(ln 2x) 3. f (x) = x 4. f (x) = 3 By the Chain rule, therefore, f (x) = 1 4x 4x + 2 2 1 + 2x 1 - 2x2 (1 - 2x2 ) + (1 + 2x2 ) 1 - 4x2 . sin(ln 2 x) x 2 sin(ln 2x) 5. f (x) = - x 6. f (x) = - sin(ln 2 x) Explanation: By the Chain Rule sin(ln 2x) f (x) = - . x = 2x Consequently, f (x) = 4x 1 - 4x4 . 006 10.0 points Find the value of f (e7 ) when f (x) 1. = f (e7 ) = 10 49 10 49 x . ln x 005 10.0 points Find the derivative of f (x) = ln 1. f (x) = - 2. f (x) = - 3. f (x) = 4. f (x) = 4x 1 - 4x4 2x3 1 - 4x2 4x3 1 - 4x4 4x correct 1 - 4x4 4x3 1 - 4x4 1 + 2x2 . 1 - 2x2 2. f (e7 ) = - 3. f (e7 ) = 4. f (e7 ) = 13 49 6 correct 49 6 49 13 49 5. f (e7 ) = - 6. f (e7 ) = - 5. f (x) = - 6. f (x) = 2x 1 - 4x2 Explanation: By the Quotient Rule, f (x) = Consequently, ln x - x(1/x) ln x - 1 = . (ln x)2 (ln x)2 Explanation: Properties of logs ensure that ln 1 + 2x2 1 - 2x2 = ln = (1 + 2x2 ) - ln (1 - 2x2 ) f (e7 ) = since ln(e7 ) = 7. 007 6 49 1 ln(1 + 2x2 ) - ln(1 - 2x2 ) . 2 10.0 points whang (cw27622) HW 12 - Implicit and Log Differentiation bormashenko (55335) Determine f (x) when f (x) = e(2 ln(x 1. f (x) = 9x10 2. f (x) = 1 2 ln(x5 ) e x 5 5 4 )) . 3. f (x) = 10(ln x)e2 ln(x 4. f (x) = 2 2 ln(x5 ) e x2 ) x4y = y 5x 4y ln x = 5x ln y 1 1 4y + 4(ln x) y = 5x y + 5(ln y) x y 1 1 4y ln x - 5x y = 5(ln y) - 4y y x y(5x ln y - 4y) y = x(4y ln x - 5x) 009 10.0 points 5. f (x) = 10x9 correct 6. f (x) = e10/x Explanation: Since r ln x = ln xr , we see that f (x) = e(ln x Consequently, 10 Determine y when y = xx . 1. y = y(ln x + 1) correct eln x = x , 2. y = -xy(2 ln x + 1) 3. y = -y(ln x + 1) ) = x10 . 4. y = -y(2 ln x - 1) 5. y = xy(2 ln x - 1) f (x) = 10x9 . 008 10.0 points Find y if x = y 5x . 4y y (ln x - 1) x2 y 7. y = - 2 (ln x - 1) x 6. y = 8. y = xy(2 ln x + 1) Explanation: After taking natural logs we see that ln y = x ln x . Thus by implicit differentiation, 1 dy = ln x + 1 . y dx Consequently, y = y(ln x + 1) . 1. y = 4y 2 (ln x + 1) 4y ln x - 5x 4y 2 (ln x - 1) 2. y = 4y ln x + 5x y 2 (ln x - 1) 3. y = 4y ln x - 5x 4. y = 5. y = y(5x ln y - 4y) correct x(4y ln x - 5x) y(4x2 ln y - 5y) x(5y 2 ln x - 4x) Explanation: whang (cw27622) HW 12 - Implicit and Log Differentiation bormashenko (55335) 010 10.0 points 5 Find the value of f (-1) when f (x) = 3 tan-1 x - 6 sin-1 x . 1. f (-1) = 2. f (-1) = 3. f (-1) = 11 4 9 correct 4 15 4 Explanation: Differentiating implicitly with respect to x we see that 1 dy y+x +1 = 0 , xy dx in which case y(1 + x) e-x (1 + x) dy = - = - dx x x2 because, by exponentiation, e-x . y = x Consequently, at x = 1, slope = 012 dy dx x=1 = -2e-1 . 17 4. f (-1) = 4 5. f (-1) = Explanation: Since tan-1 (-1) = - , 4 we see that f (-1) = 3 9 3- = . 4 4 sin-1 (-1) = - , 2 13 4 10.0 points Find the derivative of 1 x tan-1 f (x) = 3 3 x 2 tan-1 1. f (x) = 2 1 + 9x 3 2. f (x) = 3. f (x) = 4. f (x) = 1 x tan-1 2 9+x 3 2 . 2 x correct tan-1 9 + x2 3 011 10.0 points Find the slope of the line tangent to the graph of ln(xy) + x = 0 at the point where x = 1. 1. slope = -e 2. slope = e 3. slope = 2e-1 4. slope = -2e-1 correct 5. slope = e-1 6. slope = -2e 1 x x tan sec2 3 3 3 x x tan 5. f (x) = 2 sec2 3 3 6. f (x) = 1 x tan-1 2 1 + 9x 3 Explanation: Since d 1 tan-1 x = , dx 1 + x2 the Chain Rule gives d 3 x = tan-1 . dx 3 9 + x2 Using the Chain Rule yet again, therefore, we see that 2 x f (x) = . tan-1 2 9+x 3 keywords: derivative, inverse tan, Chain Rule,
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