dx dx [x2 + y 2 ] = y d [tan x] + tan x = 2x + 2y = y sec 2 x + tan x dy dy = tan x dy dy = [tan x 2y] dy dx = 2x y sec2 x [1 + sin y] = sin(xy)

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1 Math 7 Activit: Implicit & Logarithmic Differentiation (Solutions) Implicit Differentiation. For each of the following equations, etermine x. a. tan x = x tan x] = x x x2 + 2 ] = tan x] + tan x = 2x x x x = sec 2 x + tan x = 2x x x = tan x 2 x x = 2x sec2 x = tan x 2] x = 2x sec2 x = x = 2x sec2 x tan x 2 b. cos(x) = + sin x cos(x)] = + sin ] = sin(x) x x = sin(x) x ] x + x] = 0 + cos x = cos x = x sin(x) sin(x) = cos x x = sin(x) = x sin(x) + cos x x = sin(x) = x sin(x) + cos ] x = x = sin(x) x sin(x) + cos

2 c. e cos x = + sin(x) x e cos x] = + sin(x)] = e cos x] + cos x x x x e ] = 0 + cos(x) x x] x ] x + = e sin x + cos x e x = cos(x) = e sin x + e cos x = x cos(x) x x + cos(x) = e cos x x cos(x) x x = e sin x + cos(x) = e cos x x cos(x)] x = e sin x + cos(x) = x = e sin x + cos(x) e cos x x cos(x) 2. Fin an equation of the tangent line of the curve e x = 6 at the point (2, 2). We first etermine x : 2 x ex ] = 6] = 2 x x ex 4 ] + e x2 4 x x = 0 = 2xe x2 4 ] + e x2 4 x x = 0 = e x2 4 x x = 2 2xex 4 = e x2 4 ] x = 2 2xex 4 = x = 2xex 2 4 e x2 4 Therefore, x = 2(2)(2)e0 (2,2) e 0 + 2(2) = 8 5 Hence, an equation of the tangent line to the given curve at the point (2, 2) is 2 = 8 (x 2) 5 or = 8 5 x

3 3. Determine if there are an points on the curve x x = 2 whose tangent lines are horizontal. Fin each point (if there are an). We begin b ifferentiating both sies of the given equation with respect to x: x x2 2 + x] = ( 2] = x2 2 ) + 2 (2x) + x x x x + = 0 () Since we are intereste in fining points on the curve whose tangent lines have slope zero, we set = 0. Therefore, from (), we solve the following equation: x 2x 2 + = 0 = (2x + ) = 0 = = 0 or x = 2 Next, we plug these two results into the given equation. Note that plugging = 0 into the given equation results in the equation 0 = 2, which means that = 0 is an extraneous solution, i.e. there is no horizontal tangent line at an point whose -coorinate is zero. Next, we substitute x = 2 into the given equation to obtain ( 2 ( + 2) ) = 2 (2) 2 which is false since the left-han sie of (2) is equal to 4. Therefore, we conclue that there is no point on the given curve that has a horizontal tangent line. 4. Two curves are orthogonal if their tangent lines are perpenicular at each point of intersection. Show that the following families of curves are orthogonal trajectories of each other; that is, ever curve in one famil is orthogonal to ever curve in the other famil. x = c, x 2 2 = k (where c an k are constants.) Recall that two nonvertical lines are perpenicular to each other if the slope of one of the lines is the negative reciprocal of the slope of the other. We want to show that the tangent lines of both families of curves at their points of intersection are alwas perpenicular. Note that the two equations each represent a famil of hperbolas. If (x 0, 0 ) is a point of intersection, then x 0 0 = c an x = k. We now etermine x for both curves. For the first famil of hperbolas, we have x x] = c] = x x x + = 0 = x = x Therefore, the slope of the tangent line to x = c at the point (x 0, 0 ) is 0 x 0. For the secon famil of hperbolas, we have x x2 + 2 = k] = 2x 2 x = 0 = x = x Therefore, the slope of the tangent line to x 2 2 = k at the point (x 0, 0 ) is x 0 0, which is the negative reciprocal of 0 x 0. Hence, the curves are orthogonal, an the families of curves are orthogonal trajectories of each other.

4 Logarithmic Differentiation 5. Given = (cos x) sin x, compute x. We begin b appling the natural logarithm on both sies of the given equation: B the properties of logarithms, we can write (3) as Differentiating both sies of (4) an appling the Chain Rule iels which can be simplifie to Next, we isolate x ln = ln (cos x) sin x] (3) ln = sin x ln(cos x) (4) x = sin x ( sin x) + cos x ln(cos x) (5) cos x Since = (cos x) sin x, (7) becomes = sin x tan x + cos x ln(cos x) (6) x b multipling both sies of (6) b : x = sin x tan x + cos x ln(cos x) ] (7) x = (cos x)sin x sin x tan x + cos x ln(cos x) ] 6. Let = x 2 cos(2x) (2x 2 + ) 0 sin(3x). We begin b appling the natural logarithm on both sies of the given equation: x 2 ] cos(2x) ln = ln (2x 2 + ) 0 sin(3x) (8) B the propert of logarithms of quotients, (8) can be rewritten as ln = ln x 2 cos(2x) ] ln (2x 2 + ) 0 sin(3x) ] (9) Furthermore, b the propert of logarithms of proucts, (9) can be rewritten as ln = ln(x 2 ) + ln cos(2x) ] {ln (2x 2 + ) 0] + ln sin(3x) ] } (0) = ln(x 2 ) + ln cos(2x) ] ln (2x 2 + ) 0] ln sin(3x) ] () Since ln(x 2 ) = 2 ln x an ln (2x 2 + ) 0] = 0 ln(2x 2 + ), we have ln = 2 ln x + ln cos(2x) ] 0 ln(2x 2 + ) ln sin(3x) ] (2)

5 Differentiating both sies of (2) an appling the Chain Rule iels which can be simplifie as x = 2 x + 0 ( 2 sin(2x)) cos(2x) 2x 2 + (4x) (3 cos(3x)) (3) sin(3x) x = 2 x 2 tan(2x) 2x 2 3 cot(3x) (4) + Multipling both sies of (4) b iels Since = x = 2 x 2 tan(2x) x 2 cos(2x) (2x 2 + ) 0, (5) becomes sin(3x) x = x 2 cos(2x) 2 (2x 2 + ) 0 sin(3x) x 2 tan(2x) ] 2x cot(3x) ] 2x cot(3x) (5) (6)