(P+2) (V2z+2Ar+1)(v2x+1)[/2x+Iv/2z+2Az+1] Oh (t) -8t3 -3 (t² +2)* (V2r+2Az+1)(v2zr+1)[/2x+Iv2z+2Az+I] (16+4*) O (E) None of the choices O (E) None of the choices (16:4") 8t3 O h' (t) = (1² +2)® -3 (/2x+2Az+1)(/2x+1)[/2x+I+v2z+2Az+1] O (E) None of the choices 2. In the Four-Step Rule, which among the 5. Given: y = ; Find y'. steps is the most important, upon which 3(x2 –3x+5) 8. Given y=[ 2 +(x²+1)ª ]³; Find y' the foundation of determining differential equations rests? 5- (x2 – 3.x+5)² y' = 48x (x²+1)³ [ 2 +(x²+1)^ ]² O Step 2 (subtract f(x) from f(x + Ax) and simplify) 2² –5 O y'=24x (x2+1)³ [2 +(x²+1)ª ]² -3r+5) O Step 4 (apply the limit Ax → 0) and simplify. O (E) None of the choices O (E) None of the choices O (E) None of the choices 2² -5 O Step 1 (write f(x) as f(x + Ax)) (2² –3r+5)? O y' = 4 (x2+1)° O Step 3 (divide the difference between f(x + Ax) and f(x) by Ax) 5-2? (x² – 3æ+5)² O y' = 64 (x2+1)° 3. Which of the following statements is true? 6. Find dy/dx, given the following: y = (2ª +27)° 9. Given y = Vx – 2/x – 2; Find y'. V2x-4 O (E) None of the choices VE-2–1 (z²+2=)? V(2z-4) 3(3z²+2) (z²+2z) V2z-4 O y = O y = 2/(w-2)(2–2,#-2) If a curve y=f(x) has a horizontal tangent line at X=a, then f'(a) is undefined. O (E) None of the choices O (E) None of the choices (2* +2z)| 17z–36z² +10z–24 V(2z–4) The slope of the tangent line to the curve y = O y = O y = 1-VI-2 (2z–4)? x^2 -2x at point (2,0) is 4. 2(x-2)7 3(32² +2) (z" +2z)* V(2z-4) (z*+2z)² O y = - 1-Va-2 If a curve y=f(x) has a vertical tangent line at 2/(x-2)(x-2/æ–2) x=a, then f'(a)=0. O- (2°+2z)“ ( 17z³ – 36x²+10z–24 V2z-4 22-4 VE-2-1 2(x-2) 7 If the tangent line to the graph of y=f(x) at x-3 has a negative slope, then f'(3) < 0. 10. Find the slope of the line that is tangent to the graph of f(x)=3x²-2 at P(-3, 25). O (E) None of the choices O -3/25 O -18 O 27
(P+2) (V2z+2Ar+1)(v2x+1)[/2x+Iv/2z+2Az+1] Oh (t) -8t3 -3 (t² +2)* (V2r+2Az+1)(v2zr+1)[/2x+Iv2z+2Az+I] (16+4*) O (E) None of the choices O (E) None of the choices (16:4") 8t3 O h' (t) = (1² +2)® -3 (/2x+2Az+1)(/2x+1)[/2x+I+v2z+2Az+1] O (E) None of the choices 2. In the Four-Step Rule, which among the 5. Given: y = ; Find y'. steps is the most important, upon which 3(x2 –3x+5) 8. Given y=[ 2 +(x²+1)ª ]³; Find y' the foundation of determining differential equations rests? 5- (x2 – 3.x+5)² y' = 48x (x²+1)³ [ 2 +(x²+1)^ ]² O Step 2 (subtract f(x) from f(x + Ax) and simplify) 2² –5 O y'=24x (x2+1)³ [2 +(x²+1)ª ]² -3r+5) O Step 4 (apply the limit Ax → 0) and simplify. O (E) None of the choices O (E) None of the choices O (E) None of the choices 2² -5 O Step 1 (write f(x) as f(x + Ax)) (2² –3r+5)? O y' = 4 (x2+1)° O Step 3 (divide the difference between f(x + Ax) and f(x) by Ax) 5-2? (x² – 3æ+5)² O y' = 64 (x2+1)° 3. Which of the following statements is true? 6. Find dy/dx, given the following: y = (2ª +27)° 9. Given y = Vx – 2/x – 2; Find y'. V2x-4 O (E) None of the choices VE-2–1 (z²+2=)? V(2z-4) 3(3z²+2) (z²+2z) V2z-4 O y = O y = 2/(w-2)(2–2,#-2) If a curve y=f(x) has a horizontal tangent line at X=a, then f'(a) is undefined. O (E) None of the choices O (E) None of the choices (2* +2z)| 17z–36z² +10z–24 V(2z–4) The slope of the tangent line to the curve y = O y = O y = 1-VI-2 (2z–4)? x^2 -2x at point (2,0) is 4. 2(x-2)7 3(32² +2) (z" +2z)* V(2z-4) (z*+2z)² O y = - 1-Va-2 If a curve y=f(x) has a vertical tangent line at 2/(x-2)(x-2/æ–2) x=a, then f'(a)=0. O- (2°+2z)“ ( 17z³ – 36x²+10z–24 V2z-4 22-4 VE-2-1 2(x-2) 7 If the tangent line to the graph of y=f(x) at x-3 has a negative slope, then f'(3) < 0. 10. Find the slope of the line that is tangent to the graph of f(x)=3x²-2 at P(-3, 25). O (E) None of the choices O -3/25 O -18 O 27
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Kindly answer the following and show the complete solutions.
![3
2
7. Given: y =
; Find: dy/dt
1. Given: h (t) = :Fir
; Find h'(t).
4. Using the four-step procedure in finding the
V16+tª
t² +2
3
what will be the equation
derivative of
V2x+1'
before step 4?
-6t3
8t° (1+t*)
(1² +2)³
Oh (t)
(16+t')
(V2.r+2Ar+1)(/2x+1)[/2z+I+v2x+2Az+1]
-2t3
h' (t)
8t° (t° –1)
(t² +2)*
(16+t*) 2
-6
(v/2z+2Az+1)(v2z+1)[/2z+Iv2x+2Az+1]
Oh' (t)
-813
-3
(1² +2)*
(/2r+2Az+1)(v2x+1)[/2x+Iy/2x+2Az+1]
(16+t“) 7
O (E) None of the choices
O (E) None of the choices
Oh' (t)
(16+41) 을
8t3
-3
(t² +2)°
(V2x+2Ax+1)(/2x+1)[/2x+I+v2x+2Ax+1]
O (E) None of the choices
2. In the Four-Step Rule, which among the
5. Given: y =
-; Find y'.
3(x2 – 3x+5)
steps is the most important, upon which
8. Given y=[ 2 +(x²+1)ª ]³; Find y'
the foundation of determining
differential equations rests?
5-z
(22 – 3.x+5)²
y' = 48x (x²+1)³ [ 2 +(x²+1)ª ]²
Step 2 (subtract f(x) from f(x + Ax) and
simplify)
2² -5
(2² –-3x+5),
y'=24x (x²+1)³ [ 2 +(x²+1)ª ]²
O Step 4 (apply the limit Ax → 0) and
simplify.
(E) None of the choices
O (E) None of the choices
O (E) None of the choices
a -5
(22 –3.0+5)²
O Step 1 (write f(x) as f(x + Ax))
O y' = 4 (x2+1)°
O Step 3 (divide the difference between f(x
+ Ax) and f(x) by Ax)
5-a?
(x² – 3x+5)*
O y' = 64 (x2+1)°
3. Which of the following statements is true?
(2² +2x)³
V2x-4
2Vx – 2; Find y'.
6. Find dy/dx, given the following: y =
9. Given y = Vx
O (E) None of the choices
Vz-2–1
(2*+2x)²
V (2z-4)
3(32° +2) (z*+2z)
V2z-4
O y =
O y =
2/(-2)(2-2,#-2)
If a curve y=f(x) has a horizontal tangent line at
x=a, then f'la) is undefined.
O (E) None of the choices
O (E) None of the choices
(23 +2z)| 17z _36z² +10x–24
V(2z–4)
The slope of the tangent line to the curve y =
O v =
O y =
1-V7-2
(2x-4)?
x^2 -2x at point (2,0) is 4.
2(x-2)7
3(3z² +2) (z* +2z)ª
O y =
1-VI-2
V (2z-4)=
If a curve y=f(x) has a vertical tangent line at
2/ (a-2)(x-2/¤-2)
x=a, then f'(a)=0.
O y =
(z²+2z)ª
17 -36x +10zr -– 24
Vz-2-1
V2z-4
2x-4
2(x–2) 7
If the tangent line to the graph of y=f(x) at x=3
has a negative slope, then f'(3) < 0.
10. Find the slope of the line that is tangent to the
graph of f(x)=3x²-2 at P(-3, 25).
O (E) None of the choices
O -3/25
O -18
O 27](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47c03568-fe9d-4d23-a941-933e6e6a6ddb%2Fe099c374-8aab-4de3-a041-dbd818b6def0%2Fplm9oj_processed.png&w=3840&q=75)
Transcribed Image Text:3
2
7. Given: y =
; Find: dy/dt
1. Given: h (t) = :Fir
; Find h'(t).
4. Using the four-step procedure in finding the
V16+tª
t² +2
3
what will be the equation
derivative of
V2x+1'
before step 4?
-6t3
8t° (1+t*)
(1² +2)³
Oh (t)
(16+t')
(V2.r+2Ar+1)(/2x+1)[/2z+I+v2x+2Az+1]
-2t3
h' (t)
8t° (t° –1)
(t² +2)*
(16+t*) 2
-6
(v/2z+2Az+1)(v2z+1)[/2z+Iv2x+2Az+1]
Oh' (t)
-813
-3
(1² +2)*
(/2r+2Az+1)(v2x+1)[/2x+Iy/2x+2Az+1]
(16+t“) 7
O (E) None of the choices
O (E) None of the choices
Oh' (t)
(16+41) 을
8t3
-3
(t² +2)°
(V2x+2Ax+1)(/2x+1)[/2x+I+v2x+2Ax+1]
O (E) None of the choices
2. In the Four-Step Rule, which among the
5. Given: y =
-; Find y'.
3(x2 – 3x+5)
steps is the most important, upon which
8. Given y=[ 2 +(x²+1)ª ]³; Find y'
the foundation of determining
differential equations rests?
5-z
(22 – 3.x+5)²
y' = 48x (x²+1)³ [ 2 +(x²+1)ª ]²
Step 2 (subtract f(x) from f(x + Ax) and
simplify)
2² -5
(2² –-3x+5),
y'=24x (x²+1)³ [ 2 +(x²+1)ª ]²
O Step 4 (apply the limit Ax → 0) and
simplify.
(E) None of the choices
O (E) None of the choices
O (E) None of the choices
a -5
(22 –3.0+5)²
O Step 1 (write f(x) as f(x + Ax))
O y' = 4 (x2+1)°
O Step 3 (divide the difference between f(x
+ Ax) and f(x) by Ax)
5-a?
(x² – 3x+5)*
O y' = 64 (x2+1)°
3. Which of the following statements is true?
(2² +2x)³
V2x-4
2Vx – 2; Find y'.
6. Find dy/dx, given the following: y =
9. Given y = Vx
O (E) None of the choices
Vz-2–1
(2*+2x)²
V (2z-4)
3(32° +2) (z*+2z)
V2z-4
O y =
O y =
2/(-2)(2-2,#-2)
If a curve y=f(x) has a horizontal tangent line at
x=a, then f'la) is undefined.
O (E) None of the choices
O (E) None of the choices
(23 +2z)| 17z _36z² +10x–24
V(2z–4)
The slope of the tangent line to the curve y =
O v =
O y =
1-V7-2
(2x-4)?
x^2 -2x at point (2,0) is 4.
2(x-2)7
3(3z² +2) (z* +2z)ª
O y =
1-VI-2
V (2z-4)=
If a curve y=f(x) has a vertical tangent line at
2/ (a-2)(x-2/¤-2)
x=a, then f'(a)=0.
O y =
(z²+2z)ª
17 -36x +10zr -– 24
Vz-2-1
V2z-4
2x-4
2(x–2) 7
If the tangent line to the graph of y=f(x) at x=3
has a negative slope, then f'(3) < 0.
10. Find the slope of the line that is tangent to the
graph of f(x)=3x²-2 at P(-3, 25).
O (E) None of the choices
O -3/25
O -18
O 27
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