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WorksheetsDE Diagnostic Test
Total questions: 30
Worksheet time: 15mins
dxd(2 sin 3x)
2 cos 3x
3 cos 3x
6 cos 3x
−6 cos 3x
dxd(2x3− x24)
6x2−x8
2x2−x4
6x2+x8
3x2+x4
Find dy/dx:
y=sin (ln x2)2 cos ln x2
x2 cos (ln x2)
2x cos (ln x2)
2 cos (ln x2)
Find dy/dx:
y=sin (ln x2)2 cos ln x2
x2 cos (ln x2)
2x cos (ln x2)
2 cos (ln x2)
If the first derivative of a function is a constant, then the function is
linear
quadratic
exponential
cunic
dxd(ln cos x)
sec x
−tan x
−sec x
tan x
dxd(tan 4x)
4 sec 4x
4 sec24x
sec24x
4 sec4x tan 4x
dxd(4x+1)
24x+1
44x+1
4x+14
4x+12
dxd(2e−0.5x)
e−0.5x
−e−0.5x
2e−0.5x
−2e−0.5x
dxd(tan−15x)
1+25x25
1+5x21
1+25x21
1+5x25
Given that
x=t2+2t and y=2t3−6t , Find dxdy2t+23t2−6
(2t+2)(6t2−6)
2t+26t2−6
6t2−62t+2
dxd(x2−42x)
x2−42x2
−(x2−4)38
−x2−48
(x2−4)32x2−4
∫e−2xdx
2ex+C
−2e−2x+C
21e−2x+C
−21e−2x+C
∫ 4−xdx
ln (4−x)+C
−ln(4−x)+C
41ln(4−x)+C
21ln(4−x)+C
∫xex dx
xex−1+C
ex(x−1)+C
xex−x+C
ex−x+C
∫ln x dx
x ln x−x +C
x ln x − 1 +C
ln x − 1+C
x ln x+1+C
∫6e3xdx
2e3x+C
31e3x+C
21e3x+C
3e3x+C
∫ 9+4x2dx
21tan−1(32x)+C
31tan−1(32x)+C
61tan−1(32x)+C
31tan−1(92x)+C
∫ 4−x2dx
sin−1(2x)+C
21sin−1x+C
sin−1x+C
sin−1(4x)+C
∫sec t dt
secttant+C
sect+tant+C
ln∣sectant∣+C
ln∣sect+tant∣+C
∫ 2x+14dx
22x+1+C
42x+1+C
412x+1+C
212x+1+C
∫cscx cotx dx
csc x +C
−csc x+C
cot x+C
−cotx+C
∫csc2x dx
−cotx+C
−cotxcscx+C
cotx+C
cotxcscx+C
∫cos 2x dx
21sin2x+C
2 sin 2x+C
−21sin2x+C
−2 sin 2x+C
∫(2x+5)3dx
41(2x+5)4+C
81(2x+5)4+C
21(2x+5)4+C
(2x+5)4+C
∫20x3 dx
4x4+C
5x4+C
10x4+C
20x4+C
∫cos2 t sint dt
31cos3t+C
−31cos3t+C
3cos3t+C
−3cos3t+C
∫sin 2x dx
21cos 2x+C
−21cos 2x+C
2cos 2x+C
−2cos 2x+C
What is the Integration by Parts Formula?
What would you choose for your u here if you used integration by parts?
t
3t
e2t
et
don't use IBP, let u = 2t
