Differential Calculus

556: CE Board November 1997 Evaluate:

lim

x→ 1

x 2−1 x 2+3 x−4

A. 1/5 B. 2/5 C. 3/5 D. 4/5 Solution:

lim

x→ 1

x 2−1 x 2+3 x−4

Substitute x=1:

(1)2 −1 0 = , indeterminate 2 0 x→ 1 (1) +3(1)−4

lim

Note: Using L’Hospital’s rule, differentiate separately the numerator and denominator and substitute the value of limit to the variable.

lim

2x 2 x +3

lim

2(1) 2 = 2 (1 )+3 5

x→ 1

x→ 1

557: ECE Board April 1998 Evaluate:

(x−4) x → 4 ( x −x−12) lim

A. Undefined B. 0 C. Infinity D. 1/7 Solution:

2

(x−4) x → 4 (x −x−12) lim

2

Substitute x=4:

(4−4 ) 0 = 0 , Indeterminate x → 4 (4 −4−12) lim

2

Apply L’Hospital’s rule & substitute x=4:

lim

1 2 x −1

lim

1 1 = 2( 4)−1 7

x→4

x→4

558. ME Board April 1998 Evaluate:

lim

x→4

x 2−16 x−4

A. 0 B. 1 C. 8 D. 16 Solution:

lim

x→4

x 2−16 x−4

Substitute x=4:

lim

x→4

4 2−16 0 = 4−4 0 , Indeterminate

Apply L’Hospital’s rule & substitute x=4:

lim

x→4

2x =2 x 1

lim 2 ( 4 )=8 x→4

559: ECE Board April 1993

M =lim

Evaluate:

x →2

x 2−4 x−2

A. 0 B. 2 C. 4 D. 6 Solution: 2

lim

x→ 2

x −4 x−2

Substitute x=2:

lim

x→ 2

22−4 0 = 2−2 0 , Indeterminate

Apply L’Hospital’s rule & substitute x=2:

lim

x→4

2x =2 x 1

lim 2 ( 2 )=4 x→4

560: EE Board April 1995 Evaluate: A. 0 B. ½ C. 2 D. -1/2 Solution:

lim

x→ 0

1−cos x x2

lim

1−cos x 2 x

lim

1−cos 0 0 = 2 0 , Indeterminate 0

x→ 0

x→ 0

Apply L’Hospital’s rule & substitute x=0:

lim

sinx 2x

lim

sin 0 0 = 2(0) 0 , Indeterminate

x→ 0

x→ 0

Apply L’Hospital’s rule again:

lim

cosx 2

lim

cos 0 1 = 2 2

x→ 0

x→ 0

561: ME Board October 1997 Compute the following limit:

lim

x→∞

x+ 4 x −4

A. 1 B. 0 C. 2 D. Infinite Solution:

lim

x→∞

x+ 4 x −4

Substitute x=∞:

lim

x→∞

∞ +4 ∞ = ∞−4 ∞ , Indeterminate,

Apply L’Hospital’s rule & substitute x=∞:

1 lim =1 x→∞ 1

562: EE Board October 1994

3 x 4 −2 x 2 +7 lim Evaluate: x → ∞ 5 x 3+ x −3 A. Undefined B. 3/5 C. Infinity D. Zero Solution:

lim

x→∞

3 x 4 −2 x 2 +7 5 x 3+ x −3

Substitute x=∞:

3 ∞ 4−2 ∞ 2 +7 ∞ lim = 3 ∞ , Indeterminate x → ∞ 5 ∞ +∞−3 Apply L’Hospital’s rule and x=∞:

12 x 3−4 x 2 x → ∞ 15 x + 1 lim

12 ∞ 3−4 ∞ ∞ lim = 2 ∞ , Indeterminate x → ∞ 15 ∞ + 1 Apply L’Hospital’s rule again:

lim

36 x 2−4 30 x

lim

36 ∞ 2−4 ∞ = 30 ∞ ∞ , Indeterminate

x→∞

x→∞

Apply again L’Hospital’s rule:

72 x 72 ∞ = =∞ 30 x → ∞ 30 lim

563: EE Board November 1991 Evaluate:

lim ( x 2 +3 x−4 ) x →4

A. 24 B. 26 C. 28 D. 30 Solution:

lim ( x2 +3 x−4 ) ⁡x→ 4

Substitute x=4:

lim ( 4 2+3 (4)−4 ) x→4

lim ¿ 24 x→4

564: ECE Board November 1994 tan

Evaluate: A.

e2π

B.

e

2/ π

C. 0 D.

∞

Solution:

lim (2−x ) x→ 1

πx 2

tan

lim (2−x )

πx 2

x→ 1

Substitute x=1: tan

lim (2−1)

π (1) ∞ =1 2

x→ 1

, Indeterminate

Take in both sides: tan

In

lim ln (2−x )

In

(2−x ) ln ¿ πx lim tan ¿ 2 x→ 1

¿

In

πx 2

x→ 1

ln(2−x ) 1 πx tan 2 lim x→ 1

ln( 2−x) πx cot 2

Note: D (lnu) =

du u

and d (tan u) = -csc2 u du

Apply L’Hospital’s rule:

−1 −1 2−x 2−1 lim = In x→ 1 πx π cos 2 ( ) −csc 2 π (1 ) ( π ) 2 2 2 2 ¿

−1 π = −π 2 2

In Lim =2/π Take exponential on both sides:

eln Lim=e2/π Lim=e2/π 565: EE Board October 1997 Differentiate

x

y=e cos x

A.

– e x sin x 2 x

B.

x 2−2 x sin x 2 cos ¿ ex ¿

C.

e x cos x 2−2 x sin x2

D.

−2 xe sin x

2

x

Solution: y=ex cos x2 Note: d (uv) = udv + vdu where: u=ex; du=ex; v=cos x2; dv=-2x sinx2 y’=ex (-2x sinx2) + cos x2ex y’=ex (cosx2-2x sin x2)

566: EE Board October 1997 Differentiate

y=sec (x 2+ 2)

A.

2 x cos( x2 +2)

B.

−cos ( x 2+2 ) cot (x 2+2)

C.

2 x sec(x 2 +2) tan(x 2+ 2) 2

D. cos( x +2)

Solution: y=sec(x2+2) Note: d sec u=sec u tan u du Where: u=x2 + 2; du=2x y’=sec(x2+2) tan(x2+2) (2x) y’=2xsec(x2+2) tan(x2+2) 567: CE Board November 1994 3

What is the derivative with respect to x of (x+ 1) −x A. 3x+6 B. 3x-3 C. 6x-3 D. 6x+3 Solution: y=(x+1)3-x3 Note: d (u) n=nun-1du y’=3(x+1)2-3(x) 2(1) y’=3(x+1)2-3(x) 2 y’=3(x2+2x+1)-3x2 y’=3x2+6x+3-3x2 y’=6x+3

3

?

568: EE Board October 1997 Differentiate

y=log 10( x 2 +1)2

x 2+1 4 x¿

A.

4 x log 10 e

B.

2

x +1

C.

log e ( x ) ( x 2+ 1)

D.

2 x (x 2 +1)

Solution:

y=log 10( x 2 +1)2 du Note: d (log10u) =log10e ( u ) where: u=(x2+1)2 du=2(x2+1) (2x) =4x(x2+1)

[

4 x( x 2 +1) y =log 10 e ( x 2 +1)2 '

y'=

]

4 x log 10 e x 2+ 1

569: EE Board October 1997 Differentiate 2

A.

1 /2

(x + 2) 2

(x 2+ 2)1 /2

B.

x 1 /2 ( x + 2)

C.

2x 1 /2 ( x + 2)

D.

(x 2+ 2)3 /2

2

2

Solution: 2

1 /2

y= (x + 2)

Note: d (u) n=nun-1du Where: n=1/2; u=x2+2; du=2x 1

−1 1 2 2 ( x +2 ) (2 x) y’= 2

y’=x(x2+2)-1/2

x y’= ( x 2+2)1/ 2

570: EE Board October 1997 If

y=(t 2 +2)2

A.

3 2

B.

2 x +2x 3

C.

2( x+ 2)

D.

x 5/ 2+ x 1 /2

2

Solution:

and t=x

1/ 2

dy , determine dx .

t=x

1/ 2

→Eq.1

y=(t 2 +2)2 →Eq.2 Substitute Eq.1 in Eq.2:

y=[ (x 1/ 2)2+2 ]

2

y=(x+2)2 y’=2(x+2) 571: ME Board April 1997 x

What is the first derivative of the expression (xy) =e ? A. 0 B.

x y

C.

xy 1+ln ¿ ¿ ¿ −y¿

D.

xy 1−ln ¿ ¿ ¿ −y¿

Solution:

(xy) x =e Take in both sides: ln (xy) x = ln e x ln (xy) = 1 Note: d (uv) = udv + vdu Differentiating both sides:

x

[

]

'

x y +y +lnxy ( 1 )=0 xy

xy’+ y + y ln xy = 0 xy’=-y – y ln xy xy’= -y (1 + ln xy)

−y x (1 + ln xy)

y’=

572: ME Board April 1998 Find the derivative with respect to x the function

A.

−2 x2 √2−3 x 2

B.

−3 x √2−3 x 2

C.

−3 x2 √2−3 x 2

D.

3x √2−3 x 2

Solution:

y=√ 2−3 x 2 Note:

d √u=

du 2√ u

Where: u=2-3x2; du=-6x

y'=

'

y=

−6 x 2 √ 2−3 x 2 −3 x √2−3 x 2

√ 2−3 x 2

573: EE Board April 1995 Find y’ if y=arc sin cos x A. -1 B. -2 C. 1 D. 2

Solution: y=arc sin cos x Note: d (sin-1u) =

du √1−u 2

Where: u=cos x; u2=cos2x; du=-sinx

−sinx −sin x = 2 y’= √ 1−cos x √ sin 2 x −sin x sin x

y’

y’=-1 574: CE Board May 1997 Find the derivative of arc cos 4x. A.

−4 (1−16 x 2)0.5

B.

4 (1−16 x 2)0.5

C.

−4 (1−4 x 2)0.5

D.

4 2 0.5 (1−4 x )

Solution: Y=cos-1 4x

−du Note: d (cos u) = √ 1−u 2 -1

Where: u=4x; '

y=

−4 √1−16 x 2

u2=16 x 2 ; du=4

y'=

−4

( 1−16 x 2 )

0.5

575: CE Board November 1995 3

( x+ 1) x

Find the derivative of 3

3

A.

( x+ 1) (x +1) − x x

B.

4( x+1) 2(x +1) − x x

C.

2( x+1)3 ( x+1)3 − x x

D.

3 (x+1) ( x+ 1) − x x

2

3

2

3

Solution: 3

y=

(x +1) x

Note:

d

( uv )= vdu−udv v 2

Where: u=(x+1)3; du=3x(x+1)2 v=x; v2=x2; dv=1 2

y'=

3 x ( x +1)2 ( x+ 1 )3 y= − x2 x2 '

3

x ( 3 ) ( x +1) −( x+1 ) (1) 2 x

3(x+ 1)2 ( x +1 )3 y= − x x2 '

576: ECE Board November 1991 Differentiate the equation

A.

x2 +2 x ( x+ 1)2

B.

x x +1

y=

x2 x+ 1

C. 2x 2

2x x +1

D.

Solution:

y=

d

x2 x+ 1

( uv )= vdu−udv v 2

Where: u=x2; du=2x v=x + 1; v2=(x+1)2; dv=1

( x +1 ) ( 2 x )−x 2 (1) 2 x 2+ 2 x−x 2 y= = (x+ 1)2 ( x+1)2 '

y'=

2 x 2 +2 x−x 2 ( x +1)2

y'=

x 2+2 x 2 (x +1)

577: CE Board November 1995 The derivative with respect to x of 2 cos 2(x2+2) is A. 2sin (x2+2) cos (x2+2) B. -2sin (x2+2) cos (x2+2) C. 8xsin(x2+2) cos (x2+2) D. -8x sin(x2+2) cos (x2+2) Solution: y=2 cos2(x2+2)

1+cos 2θ 2

Note: cos2Θ= Simplifying: y=2cos2(x2+2) y= 2

[

1+cos 2(x2 +2) 2

]

2

y= 1+cos 2 ( x +2 ) 2

y= 1+cos( 2 x +4) Note: d (cos u) =-sin u du Where: u=2x2 + 4; du = 4x y’ = -sin (2x2+4) (4x) y’=-4sin2(x2+2) Note: sin 2Θ=2 cos Θ sin Θ y’= -4x [2cos(x2+2) sin(x2+2)] y’=-8x [cos(x2+2) sin(x2+2)] 578: CE Board November 1993 Find the second derivative of y by implicit differentiation from the equation 4x2+8y2=36. A. 64x2 B.

–

9 3 y 4

C. 32xy

−16 3 y 9

D.

Solution: 4x2+8y2=36 x2+2y2=9 → Eq. 1 Differentiate both sides: 2x + 4yy’=0 y’=

−2 x 4y −x

y’= 2 y

→ Eq. 2

Take second derivative: y’’=

[

−1 y ( 1 )−xy ' 2 y2

]

→ Eq. 3

Substitute Eq. 2 in Eq. 3:

y''=

y''=

−1 2

−1 2

y ' '=

[ ] [ ] y (1 ) −x

(−x 2y)

y2

x2 2y y2

y+

[

2

−1 2 y + x 2 2y 2y

2

]

2 y 2+ x2 y = 4 y3 ''

Substitute Eq. 1 in y’’:

y''=

−9 3 4y

579: ME Board April 1998 Find the partial derivatives with respect to x of the function xy 2-5y+6. A. y2-5 B. y2 C. xy-5y D. 2xy Solution:

∂ ( x y 2−5 y +6 ) = y 2 ( 1 ) +0 ∂x ∂ ( x y 2−5 y +6 ) = y 2 ∂x

580: ME Board October 1997 Find the second derivative of x3-5x2+x=0 A. 10x-5 B. 6x-10 C. 3x + 10 D. 3x2-5x Solution: y= x3-5x2+x y’= 3x2-10x+1 y’’=6x-10 581: ME Board April 1998 Given the function f(x) =x to the 3rd power – 6x + 2. Find the first derivative at x = 2. A. 6 B. 7

C. 3x2-5 D. 8 Solution: y=x3-6x+2 Differentiate and substitute x = 2: y’= 3x2-6 y’=3(2)2-6 y’=6 582: CE Board May 1996 Find the slope of the ellipse x2+4y2-10x-16y+5=0 at the point where y=2+80.5 and x=7. A. -0.1463 B. -0.1538 C. -0.1654 D. -0.1768 Solution: x2+4y2-10x-16y+5=0 Note: slope = y’ As given: y=2 + 80.5 = 4.828, x=7 Differentiate & substitute x = 7 & y=4.828 2x+8yy’-10-16y’=0 y’ (8y-16) =10-2x

y=

10−2 x 8 y−16

y'=

10−2(7) 8 ( 4.828 )−16

'

y’=-0.1768 583: EE Board October 1997 If y=4 cos x + sin 2x, what is the slope of the curve when x = 2 radians?

A. -2.21 B. -4.94 C. -3.25 D. 2.21 Solution: y= 4 cos x + sin 2x Differentiate: y’=4(-sinx) + cos 2x (2) y’=2cos 2x – 4sinx At x =2 radians

[ ( )] [ ( )]

y ' =2 cos 2 2

180° 180 ° −4 2 π π

y’=2cos 229.183 - 4sin 114.591 y’=-4.94 584: ECE Board November 1991 Find the slope of the line tangent to the curve y=x 3-2x+1 at x=1. A. 1 B. ½ C. 1/3 D. ¼ Solution: y=x3 -2x + 1 Note: slope = y’ Differentiate and substitute x =1: y’=3x2 – 2 y’=3(1)2 – 2 y’=1 Note: Since the line is tangent to the curve at x=1, then the slope of the line is the same as the slope of the curve at the given point. Thus, the slope of the line is equal to 1.

585: ECE Board November 1991 Give the slope of the curve at the point (1, 1): A. ¼ B. -1/4 C. 1 ¼ D. – 1 ¼ Solution:

x3 y= 4

- 2x+1

Note: slope = y’ Differentiate and substitute x=1:

1 y ' = ( 3 x 2) −2 4 1 y ' = ( 3 ) (1)2−2 4 y ' =−1

1 4

586: ECE Board November 1998 Find the slope of x2y=8 at the point (2, 2). A. 2 B. -1 C. -1/2 D. -2 Solution: x2y=8

8 y= x 2

y=

x3 4 -2x+1

Note: slope=y’ Differentiate and substitute x=2:

8(−2 x ) −16 = 3 x4 x

y’=

−16 y’= (2)3 y-=-2 587: CE Board May 1998 Find the slope of the curve x2+y2-6x+10y+5=0 at the point (1, 0). A. 1/5 B. 2/5 C. ¼ D. 2 Solution: x2+y2-6x+10y+5=0 Note: slope=y’ Differentiate and substitute x = 1 & y=0. 2x + 2yy’ – 6 + 10y’ + 0 = 0 y’ (2y + 10) = 6 – 2x

y'=

6−2 x 2 y +10

y'=

6−2(1) 4 = 2(0)+ 10 10

'

y=

2 5

588: CE Board May 1996 Find the slope of the tangent to the curve, y=2x-x 2+x3 at (0, 2). A. 1

B. 2 C. 3 D. 4 Solution: y=2x-x2+x3 Note: Slope=y’ Differentiate and substitute x=0: y’=2-2x+3x2 y’=2-2(0) +3(0)2 y’=2

589: ECE Board April 1999 Find the coordinates of the vertex of the parabola y=x 2-4x+1 by making use of the fact that at the vertex the slope of the tangent is zero. A. (2, 3) B. (3, 2) C. (-1, -3) D. (-2, -3) Solution: y=x2-4x+1 Note: slope=y’ Differentiate and y’=0: y’=2x-4 0=2x-4 x=2 Substitute x = 2 to the given equation: y= (2)2-4(2) +1=-3 Thus, the vertex is at (2, -3) 590: ECE Board April 1999 Find the equation of the normal to x2+y2=5 at the point (2, 1). A. y=2x B. x = 2y C. 2x+3y=3 D. x+y=1 Solution: x2+y2=1 Let m1=slope of the given curve Differentiate and substitute x =2 & y=1: 2x+2yy’=0 y’=

−x −2 = y 1

y’=-2 m1=-2 Note: Since the line is normal to the curve at the given point, the slope (m 2) of the line is equal the negative reciprocal of the slope of the given curve.

−1 −1 1 = = m2= m 1 −2 2 Using the point of slope form: y-y1=m(x-xz)

1

y-1= 2

( x−2)

2y-2=x-2 X=2y 591: CE Board May 1995 What is the equation of the normal to the curve x 2+y2=25 at (4, 3)? A. 5x+3y=0 B. 3x-4y=0 C. 3x+4y=0 D. 5x-3y=0 Solution: x2+y2=25 Let

m1=slope of the given curve m2= slope of the normal line

Differentiate and substitute x=4 & y=3: 2x+2yy’=0 y’=

m1=

−x −4 = y 3 −4 3 −1

m2= m 1

=

−1 3 = −4 /3 4

Using point slope form: y-y1=m(x-x1)

3

y-3= 4

(x−4)

4y-12=3x-12 3x-4y=0 592: EE Board April 1997 Locate the points of inflection of the curve y=f(x) =x 2ex. A.

−2 ± √ 3

B.

2± √ 2

C.

−2 ± √ 2

D.

2± √ 3

Solution: y=x2ex Note: d (uv) =udv+vdu d (eu) =eudu y’=x2(ex) +ex (2x) y’’=x2(ex) +ex (2x) +ex (2) +2x (ex) y’’=x2ex+4xex+2ex At point of inflection, y’’=0: 0= x2ex+4xex+2ex 0= x2+4x+2 By quadratic formula;

−4 ± √ 42−4 ( 1 ) (2) −4 ± √ 8 x= = 2 2(1) x=

−4 ± 2 √ 2 2

x=−2± √ 2 Substitute the values of x to the given equation to solve for the values of y: 2

(−2+ √ 2) =0.19 y ¿(−2+ √ 2) e

or

2

(−2− √ 2) =0.38 y ¿(−2− √ 2) e

Note: From the choices, only values of the x-coordinates are given. Thus the suggested answer is choice C. 593: ECE Board November 1991 In the curve 2+12x-x3, find the critical points. A. (2, 18) & (-2, -14) B. (2, 18) & (2, -14) C. (-2, 18) & (2, -14) D. (-2, 18) & (-2, 14) Solution: y=2+12x-x3 Note: Critical points are points wherein the slope of the curve is zero. y’=12-3x2 0=12-3x2 x2=4 x=2 Substitute the values of x to the general equation to solve for the values of y: y=2+12(2)-2(2)3=18 or y=2+12(-2)-2(-2)3 =-14 Thus, the points are (2, 18) and (-2, -14). 594: CE Board November 1997 Find the radius of curvature of a parabola y 2-4x=0 at point (4, 4). A. 22.36 units B. 25.78 units C. 20.33 units

D. 15.42 units Solution:

R=

[ 1+( y' )2 ]

3/ 2

y' '

y2-4x=0 2yy’-4=0

4

y’= 2 y

=

2 y

y ( 0 ) −2 y ' −2 y ' = 2 y’’= y2 y

2

y’’y =-2

( 2y )

4 y’’=- y 3 Substitute y=4 to solve for y’ and y’’:

2

y’= y

y’’=

2 1 = = 4 2

−4 −4 1 = 3= 3 y (4) 16

Substitute y’ and y’’ to solve for R:

[

1 2 1+( ) 2 R= 1 16

3/2

]

R=22.36 595: ECE Board November 1996 Find the radius of curvature at any point in the curve y+lncosx=0. A. cos x

B. 1.5707 C. sec x D. 1 Solution:

R=

[ 1+( y' )2 ]

3/ 2

y' '

y + ln cos x = 0 y = - ln cos x Note: d (ln u) =

du u

Where: u=cos x; du=-sin x y’=-

x =tan x ( −sin cos x )

y’’=sec2 x Substitute y’ and y’’ to solve for R:

R=

[ 1+(tan x)2 ]

3 /2

sec 2 x

Note: 1+tan2x=sec2x 3 /2

R=

R=

( sec 2 x )

sec 2 x sec 3 x 2 sec x

R=sec x