Techniques of Differentiation Selected Problems. Matthew Staley


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1 Techniques of Differentiation Selected Problems Matthew Staley September 10, 011
2 Techniques of Differentiation: Selected Problems 1. Find /dx: (a) y =4x 7 dx = d dx (4x7 ) = (7)4x 6 = 8x 6 (b) y = 1 (x4 + 7) dx = d dx (1 (x4 + 7)) = 1 (4x3 ) = x 3 (c) y = π 3 dx = d dx (π3 ) = 0 (d) y = x + (1/ ) dx = d dx ( x + (1/ )) = d dx (x)+ d dx (1/ ) = (1) + 0 = 1
3 . Find f (x): (a) f(x) =x x 7 f (x) = d dx (x 3 + x 7 ) = d dx (x 3 )+ d dx (x 7 ) =( 3)x 3 1 +( 7)x 7 1 = 3x 4 7x 8 (b) f(x) = x + 1 x f (x) = d dx (x1/ + x 1 ) = 1 x1/ 1 +( 1)x 1 1 = 1 x 1/ x = 1 x 1 x (c) f(x) = 3x 8 + x f (x) = ( 8)( 3)x 9 + = 4x x ( ) 1 x 1/ (d) f(x) =ax 3 + bx + cx + d (a, b, c, d constant) f (x) = 3ax +bx + c
4 3. Find y (1) : (a) y = y(x) = 5x 3x +1 y (x) = 10x 3 y (1) = 10(1) 3= 7 (b) y = y(x) = x3/ + x = x 1 (x 3/ + ) = x 3/ 1 +x 1 = x 1/ +x 1 ( ) 1 y (x) = x 1/ x = 1 x x y (1) = = 1 = 1 4 = 3 4. Find dx/dt: (a) x = t t dx dt = d dt (t t) = t 1 3
5 (b) x = t +1 3t = t 3t + 1 3t = 1 3 t t 1 dx dt = d dt (1 3 t t 1 ) = 1 3 +( 1)1 3 t = t 5. Find /dx x=1 for the following: (a) y = 1 + x + x + x 3 + x 4 + x 5 dx = 1 + x +3x +4x 3 +5x 4 dx = 1 + (1) + 3(1) + 4(1) 3 + 5(1) 4 x=1 = = 15 (b) y = 1+x + x + x 3 + x 4 + x 5 + x 6 x 3 = x 3 (1 + x + x + x 3 + x 4 + x 5 + x 6 ) = x 3 + x + x 1 +1+x + x + x 3 dx = 3x 4 x 3 x x +3x = 3 x 4 x x 3 +3x x dx = 3 x= (1) + 3 3(1) 1 = = 0 4
6 (c) y = (1 x)(1 + x)(1 + x )(1 + x 4 ) = (1 x )(1 + x )(1 + x 4 ) = (1 x 4 )(1 + x 4 ) =1 x 8 dx = 8x7 dx = 8(1) 7 = 8 x=1 6. Given : f(x) =x 3 3x + 1, Approximate f (1) by considering the difference quotient f(1 + h) f(1) h for values of h near 0, and then find the exact value of f (1) by differentiating. (a) Choosing h = 0.01 we obtain the following: f( ) f(1) 0.01 = f(1.01) + f(1) 0.01 = ((1.01)3 3(1.01) + 1) ((1) 3 3(1) + 1) = 0.01 = = f (1) Note that choosing h values closer and closer to zero, i.e. h =0.001, , , , will give better and better approximations to f (1). 5
7 (b) The exact value of f (1): f(x) =x 3 3x +1 f (x) = 3x 3 f (1) = 3(1) 3 =3 3 = 0 7. Find the indicated derivative: (a) d dt [16t ] = ()(16t) = 3t (b) dc, where C =πr. dr dc dr = d dr (πr) =π d dr (r) = π 8. A spherical ballon is being inflated: (a) Find the formula for the instantaneous rate of change of the volume V with respect to the radius r, given that V = 4 3 πr3. Solution: We are asked to find the change in V with respect to r, in other words, find dv dr. dv dr = (3)4 3 πr = 4πr (b) Find the rate of change of V with respect to r at the instant when the radius is r = 5. Solution: dv dr =4π(5) =4π5 = 100π r=5 6
8 9. Solve the following: (a) F ind y (0), where y =4x 4 +x 3 +3 y = 16x 3 +6x y = 48x + 1x y = 96x + 1 y (0) = 96(0) + 1 = 1 (b) F ind d4 y dx 4 x=1, where y = 6 x 4 y =6x 4 dx =( 4)6x 5 = 4x 5 ( ) d = d y dx dx dx =( 5)( 4)x 6 = 10x 6 ( ) d d y = d3 y dx dx dx 3 =( 6)10x 7 = 70x 7 ( ) d d 3 y = d4 y dx dx 3 dx 4 =( 7)( 70)x 8 = 5040x 8 d 4 y = 5040(1) 8 dx 4 x=1 = 5040 (c) Show that y = x 3 +3x +1satisfies y + xy y =0 i. First find the derivatives needed: y = x 3 +3x +1 y =3x +3 y =6x y =6 ii. Substitute: y + xy y = 6 + x(6x) (3x + 3) = 6 + 6x 6x 6 = 0 7
9 10. Find a function y = ax + bx + c whose graph has an xintercept at 1, a y intercept of , and a tangent line with a of slope of 1 at the yintercept. The yintercept at  means that when x = 0, y =, i.e. (0, ) is on the graph. (This is the easiest way to start because plugging in x=0 reduces the equation to find c immediately) y = ax + bx + c =a(0) + b(0) + c =c So y = ax + bx. Now we use the xintercept, which tells us that (1, 0) is also on the graph, therefore, 0=a(1) + b(1) 0=a + b =a + b Now we also want a tangent line with a slope of 1 at the yintercept. This means that x=0 = 1. dx y = ax + bx dx =ax + b dx = 1 = a(0) + b x=0 1 =b Substitute b = 1 into = a + b to get a = 3. y =3x x Our function is thus, 8
10 11. Find k if the curve (i) y = x + k is tangent to the line (ii) y =x. Let P be the point at (x 0,y 0 ) at which (i) and (ii) are tangent to each other. Find /dx of (i): y = x + k dx =x (iii) We need the to find x 0 so that the slopes of (iii) and (ii) are equal at (x 0,y 0 ). The slope of the (ii) is clearly : x 0 = x 0 =1 Now P is on the line y =x, we can sub x 0 = 1 and obtain y 0 =. So P = (1, ). Plug P into (i): y = x + k =1 + k 1=k 1. Find the xcoordinate of the point on the graph y = x where the tangent line is parallel to the secant line that cuts the curve at x = 1 and x =. (Try drawing a picture to help visualize this problem.) The secant line is the line that goes through the points ( 1, 1) and (, 4), (why?). The slope is then found by using the slope formula of two points: m = y y 1 x x 1 = 4 1 ( 1) = 3 3 =1. We want this to be the same slope as the tangent line of y = x which is /dx =x. Thus x =1 x = 1 is the x coordinate we are looking for. 9
11 13. Show that any two tangent lines to the parabola y = ax,a 0, intersect at a point that is on the vertical line halfway between the points of tangency. y = ax dx =ax Let P 1 =(x 1,y 1 = ax 1) and P =(x,y = ax ) be two different points of tangency. The tangent line at P 1 has the equation: y y 1 = m(x x 1 ) y ax 1 = dx (x x 1 ) x=x1 y ax 1 =ax 1 (x x 1 ) y ax 1 = (ax)x 1 ax 1 y = (ax)x 1 ax 1 Similarly, the tangent line at P is y = (ax)x ax. Set y = y and solve for x: (ax)x 1 ax 1 = (ax)x ax (ax)(x 1 x )=a(x 1 x ) x(x 1 x )=(x 1 x )(x 1 + x ) x =(x 1 + x ) x = 1 (x 1 + x ) This is the xcoordinate of a point on the vertical line halfway between P 1 and P. 10
12 14. Show that the segment of the tangent line to the graph of y = 1, x > 0, and x the coordinate axes has an area of square units. Let P 0 =(x 0,y 0 ) be a point on the curve. Then y 0 = 1 x 0 and dx = 1 x. The equation for the tangent line is: y y 0 = m(x x 0 ) y 1 = /dx x=x0 (x x 0 ) x 0 y 1 = 1 (x x x 0 x 0 ) 0 y 1 x 0 = x x x 0 y = x x 0 + x 0 This line will intersect the xaxis when y = 0. 0= x x 0 + x 0 x x 0 = x 0 x =x 0 The line will intersect the yaxis when x = 0. y = 0 x 0 + x 0 = x 0 So we have a triangle with coordinates: ( ) this triangle is: 1 x 0 (x 0 ) =. (0, x0 ), (0, 0), (x 0, 0). The area of 11
13 15. Find conditions on a, b, c, d, so that the graph of the polynomial f(x) = ax 3 + bx + cx + d has exactly one horizontal tangent. f(x) will have a horizontal tangent when f (x) =3ax +bx + c = 0. We can solve this by using the quadratic formula: x = b ± (b) 4(3a)c (3a) = b ± 4b 1ac 6a = b ± 4(b 3ac) 6a = b ± b 3ac 6a = b ± b 3ac 3a Now x will have one real solution if b 3ac = 0: ( ) b 3ac =0 b 3ac =0 b =3ac b = ± 3ac 16. Let f(x) = { x 16x, x < 9 x, x 9 Is f continuous at x = 9? Determine whether f is differentiable at x = 9. If so, find the value of the derivative there. lim f(x) = (9) 16(9) = = 63 x 9 lim f(x) = 9=3 x 9 + Since f(x) is not continuous at x = 9, it is not differentiable at x = 9. (Differentiable implies continuous, therefore not continuous implies not differentiable.) 1
14 { x 17. Let f(x) = 3 + 1, x < 1/ x, x 1/ Determine whether f is differentiable at x = 1. If so, find the value of the derivative there. ) 3 1 lim f(x) =( + 1 x 1 16 = = 3 16 lim x 1 + f(x) =3 4 ( ) 1 = 3 4 ( ) 1 = Since these are equal, f has been shown to be continuous at x = 1. (If f was not continuous then we would immediately be able to say that f is not differentiable) Now f (x) is given by: { 3x f (x) =, x < 1/ 3x. x > 1/ Now we must still show differentiability at x = 1 : ( ) 1 lim f (x) = 3 = 3 x 1 4 lim x 1 + f (x) = 3 ( ) 1 = 3 4 Therefore f is both continuous and differentiable at x = 1 with f ( ) 1 =
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