- #1
vorcil
- 398
- 0
Q1)
If (x^5)-x = 8
then
x = e^(solve for this)
not sure, x*x*x*x*x-x = 8
help please
-
q2)
ln(3x)/x^2
dy/dx = [ ]-[ ]ln(3x) / x^3
or solve for dy/dx
I get
(x^2)*(dy/dx(ln(3x))) - ln(3x)*(dy/dx(x^2)) / x^2*x^2
=(x^2)*(1/x) - ln(3x)*2x / x^4
= [x^2]-[2x]*(ln(3x)) / x^3
(is this right?)
-
q3)
if dy/dx = 3e^x
and y = 5 when x=0
then y = [ ]
I get
I think i have to integrate 3e^x
and it's just 3e^x
y = 3e^0 + c
e^0 = 1
so 3e^x + 2 = 5
y = 3e^x + 2
and I got it wrong?
-
q4)
evaluate
the integral of (1/3 * x-1)dx between 6 and 3
integrate it = 1/3*x^2/2 - x
= 1/6x^2 - x
(1/6 * 6^2 - 6 ) - (1/6 * 3^2 - 3) = 1/6*36-6 - 1/6*9-3
= 0 - 1.5-3
= -4.5
and I got it wrong
although I think during the test I may have made the mistake of doing 1.5 - 3 instead of -1.5 - 3...
thanks please mark
If (x^5)-x = 8
then
x = e^(solve for this)
not sure, x*x*x*x*x-x = 8
help please
-
q2)
ln(3x)/x^2
dy/dx = [ ]-[ ]ln(3x) / x^3
or solve for dy/dx
I get
(x^2)*(dy/dx(ln(3x))) - ln(3x)*(dy/dx(x^2)) / x^2*x^2
=(x^2)*(1/x) - ln(3x)*2x / x^4
= [x^2]-[2x]*(ln(3x)) / x^3
(is this right?)
-
q3)
if dy/dx = 3e^x
and y = 5 when x=0
then y = [ ]
I get
I think i have to integrate 3e^x
and it's just 3e^x
y = 3e^0 + c
e^0 = 1
so 3e^x + 2 = 5
y = 3e^x + 2
and I got it wrong?
-
q4)
evaluate
the integral of (1/3 * x-1)dx between 6 and 3
integrate it = 1/3*x^2/2 - x
= 1/6x^2 - x
(1/6 * 6^2 - 6 ) - (1/6 * 3^2 - 3) = 1/6*36-6 - 1/6*9-3
= 0 - 1.5-3
= -4.5
and I got it wrong
although I think during the test I may have made the mistake of doing 1.5 - 3 instead of -1.5 - 3...
thanks please mark