WorksheetsCALCULUS QUIZ-TRY-YOUR HAND
Total questions: 50
Worksheet time: 1hrs 15mins
y=2x9−x+6x65 . Find the gradient of y.
18x8−1
18x8−x−x75
18x8−1−x75
18x8−1−x55
dxd(3x)
−3 3x21
3 3x21
3 3x2
3x2
Differentiate the following with respect of x.
y=(3x+8)5
15(3x+8)6
5(3x+8)4
(3x+8)4
15(3x+8)4
y=x21+x .Find dydx .
dydx=−x32+2x1
dydx=−2x−3+21x21
dydx=−2x−3+41x−21
dydx=−2x−3
Differentiate 2xx3−4x2 with respect to x .
x2−2
x2−2x
x−2
x−2x2
Find dy2d2x for y=21x3+8x−6 .
23x2+8
3x
23x2
23x2+8x
dxd(ln (3x−4)3)
3x−49
3x−43
(3x−4)23
(3x−4)31
dxd(ln 2x−34)
−2x−32
−2x−31
2x−31
3−2x1
dxd(ln 2x−34)
−2x−32
−2x−31
2x−31
3−2x1
dxd(sin2x)
2cos2x
−2cos2x
cos2x
2cos3x
dxd (x4+6x−7)
4x3+6
3x2+5
x4+6x−7
4x5+6x2
dxd (x4+6x−7)
4x3+6
3x2+5
x4+6x−7
4x5+6x2
Differentiate with respect to x...
6x2 + 6x
6x + 2
6x + 6
6x2 + 6
Find the turning points of:
f(x)=x3+3x2-9x+1 (hint dy/dx=0)
(-3,28) and (1,-4)
(3,28) and (1,4)
(1,28) and (-3,-4)
(3,28) and (-1,-4)
What is the derivative of xn?
(n-1)xn
nxn+1
(n+1)xn-1
nxn-1
Which of the following is the indefinite integral of 2x3+7 ?
23x2
23x2+7x+c
8x4+7x+c
8x4+c
Integrate x with respect to x
x21+c
21x−21+ c
32x23+ c
23x23+ c
∫ x1+ x21 dx
x−1 + x−2+ c
x0 − x−1+ c
lnx+ x−1+ c
lnx− x−1+ c
∫5x4dx
x5
45x5+C
x5+C
20x3 + C
∫(x2−2x)dx
31x3−x2+C
x2 −2x+C
2x−2
31x3−x2
∫6x(x+2)dx
6x2+12x+C
x2 −2x+C
3x2+6x+C
2x3+6x2+C
∫(6x−1)dx
6x23+x+C
4x23−x+C
3x−21−x+C
I didn't look at my notes to see how to do this one.
∫(x21+x36)dx
−x−1−3x−2+C
−3x−3+−46x−4+C
−x−1−3x−2
−x−3x2+C
∫(5x2−7x+6)dx
35x−27x+6x+C
x+x+x+x+x+C
x3−3x2+6x+C
35x3−27x2+6x+C
∫4sin(−x)dx
4cos(x)+C
−4sin(−x)+C
4cos(−x)+C
−4cos(−x)+C
∫(5x3−16e−4x+x1)dx
2x25+4x−4x+ln∣x∣+C
5x31−4e−4x+1+C
25x23−16e−4x+ln∣x∣+C
Got lazy with fake answers
∫8x−3dx
−2x−4+C
4x−3+C
−38x−2
−4x−2+C
∫(x−1−1)dx
ln∣x∣−x+C
−2x−2−x+C
ln∣x∣−1+C
1−x+C
∫0dx
x1+C
Not Possible
x+C
C
Find the answer of ∫x2+4x dx?
X3/2+2x2
X2+4x
x2/2 +3x+C
x3 /3+2x2+C
Integrate ∫x31dx
=34x34+c
=23x32+c
=31x−32+c
=43x34+c
Integrate ∫sinx dx
=tanx+c
=−cosx+c
=−secx+c
=cosec x +c
∫2x1dx
=ln2x+c
=2 ln2x+c
=21ln2x+c
=ln2x1+c
∫tanxdx
ln∣cos∣+C
−ln∣cosx∣+C
ln∣sinx∣+c
2tan2x+c
∫(x4−ex)dx
x24−xex+C
4ln∣x∣−ex+C
4ln(x)−ex+C
4ln∣x∣+ex+C
What is the stationary point for the curve
y=x2−4A minimum at ( 0, -4)
A maximum at (0, -4)
A minimum at (0,4)
A maximum at (0,4)
Stationary point occur when....
x = 0
dxdy=0
dydx=0
dx2d2y=0
Inflection point is a nature for stationary point when...
dx2d2y<0
dx2d2y>0
dx2d2y=0
dx2d2y=0
The nature of the point is maximum, if
dy2d2x<0
dy2d2x=0
dy2d2x>0
If dy2d2x>0 , turning point is a
maximum point
minimum point
point of inflexion
y=x2−4
What is the maximum or minimum point for the curve?
minimum at ( 0, -4)
minimum at ( 0, 4)
maximum at ( 0, -4)
maximum at ( 0, 4)
∫ x2−13x+2dx
lnx2−1
21ln∣x+1∣+25ln∣x−1∣+c
23ln∣x−1∣+21ln∣x+1∣+c
25ln∣x+1∣+21ln(x−1)2+c
Break this into partial fractions.
Use the substitution u=x2 to evaluate the integral.
2x
e16−1
ex2+c
64
Choose the correct approach to solve this question.
SUBSTITUTION
BY PARTS
PARTIAL FRACTION
INVERSE
Choose the correct approach to solve this question.
SUBSTITUTION
BY PARTS
PARTIAL FRACTION
INVERSE
Find A, B, and C
A=-2, B=3, C=5
A=-9, B=1, C=0
A=1, B=1, C=6
A=-1, B=1, C=5
Set up this function as a partial fraction
(Pg. 63 in the workbooks)
∫x+22+x+56dx
∫x+27+x+53dx
∫x+23+x+57dx
∫x+26+x+52dx
