Question 4. y = 3x^4 + 3(x^2 + 5) get rid of the brackets by expanding so (3*x^2) + (3*5) = 3x^2 + 15 the equation will continue.... y= 3x^4 + 3x^2 + 15 then find dy/dx dy/dx= 12x^3 + 6x
for 5 ,y + 6 = 4x + 9y - 3x^2 , which is the correct way ,just differentiating or grouping terms..............i think the way weezy did it but not really sure?
Yes "low rider" always make y (or wat is to be differentiated) the subject of the formula before starting to differentiating the equation. Also another thing to take note of is to make put the equation in its simpliest form before differentiating....
1. y = 6x^3 + 4x^2 - 5x when doing this you multiply the coefficent by the power that value now becomes thr coefficent and 1 is minused from the power the power y=18X^2+8X-5
Here we express (2x-3)^2 as a polynomial first by expansion::::>so (2x - 3)^2 4x^2 - 12x + 9 If y 4x^2 - 12x + 9, then dy/dx = 8x - 12x^0 + 0 = 8x - 12,as x^0=1.
For all of you who tried to ans #2 you should check back your notes that miss give... I think the ans you found were wrong.
y^2 = x^3 + 16 the 1st thing 2 do is to make y the subject of the formula:
y = (x^3 + 16)^0.5 since if you make y the subject then the other side of the equation has to compensate, therefore, in this case the other side becomes the square root or to the power of 0.5
now you diff.
y = (x^3 + 16)^0.5 This is to hard to diff. so im goin to make it simpler as follows:
substitute m = x^3 + 16 dm/dx = 3x^2
rewrite: y = m^0.5 dy/dm = 0.5m^-0.5
The question asked for dy/dx. Therefore:
dy/dx = dm/dx.dy/dm dy/dx = (3x^2) X 0.5(x^3 + 16)
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ques 1
ReplyDeleteas we can see that its many terms, now the easist way to work this ques will be too Differentiate them separately
1.first you make sure there is no bracket terms
2.then make sure that there are no downstairs
3. then you look out for the police
its as simple as that
answer y = 18x^2 + 8x - 5
Question`1
ReplyDeletey=ax^n
dy/dx=(a)(n)x^n-1
so
y = 6x^3 + 4x^2 - 5x
y`=`(6)(3)x^3-1`+`(4)(2)x^2-1`-`5x
y`=`18x^2`+`8x`-`5x
quest 1
ReplyDeletefirst you look to see which one of the rules you are dealing with then you proceed.
so clearly this one is dealing with ax^n
so dy/dx = anx^n-1
so the ans. = 18x^2 + 8x - 5
Question 1
ReplyDeletethis Question deals with ax^n
therefore dy/dx = anx^n-1
so
y = 6x^3 + 4x^2 - 5x
y`=`(6)(3)x^3-1`+`(4)(2)x^2-1`-`5x
y`=`18x^2`+`8x`-`5x
4. y = 3x^4 + 3(x^2 + 5)
ReplyDeleteexpand the brackets..
y= 3x^4 + 3x^2 + 15
then find dy/dx
so dy/dx= 12x^3 + 6x
QUES 2
ReplyDeletesince i believe that the Differenal of two trems is not on the course map i would figure that u wud just have to make y the subject then Differentiate
y^2 = x^3 + 16
y = (x^3 + 16)^0.5
and well from here its simple jus follow the rules that miss gave us in class and u should be fine
answer: dy/dx = -1.5x^2(x^3 + 18)^-0.5
OR
you can Differentiate the y trem as well in this case you shall get
dy/dx = 2y dy/dx = 3x^2
then i believe u make dy/dx subject of the formula
dy/dx = 3x^2 - 2y
renjisa n if it's y^2 won't it be difficult to divided by 2.
ReplyDeletewon't it be (x^3+16)^-2?
QUESTION1
ReplyDeletey=6x^3 + 4x^2 -5x
y=18x^3-1 + 8x^2-1 -5
y=18x^2+8x-5
y = 6x^3 + 4x^2 - 5x
ReplyDelete= 18x + 8x - 5
for 2) jus square the other side trid to get rid of y^2..
ReplyDelete3) to remove the 1/y u carry across the 4x+4
ReplyDeleteso itl be 1/4x +4 = y
which is d same as (4x +4)^ -1 = y
from there you diff as normal
y = 3x^4 + 3(x^2 + 5)
ReplyDeleteremove brackets
y= 3x^4 + 3x^2 + 15
dy/dx= 12x^3 + 6x
5. y + 6 = 4x + 9y - 3x^2
ReplyDeletey -9y = 4x - 3x^2 - 6
y(-9y) =4x - 3x^2 -6
-9y^2 =4x - 3x^2 -6
y^2 = 4x - 3x^2 - 6/ -9
dy/dx= (4 - 6x)^9
dont forget u hav to square it after to get y alone...
question4
ReplyDeletey=3x^4+3(x^2+5)
y=3x^4+3x^2+15
dy/dx=12x^3+6x
question5
ReplyDeletey+6=4x+9y-3x^2
y=4x+9y-3x^2-6
y=-3x^2+4x+9y-6
y-9y=-3x^2+4x-6
-8y=-3x^2+4x-6
8y=3x^2-4x+6
y=3/8x^2-4/8x+6/8
y=6/8x-4/8
1. y = 6x^3 + 4x^2 - 5x
ReplyDeleteso you say dy/dx = 18 x ^2 + 8x -5x
you multiply by the power and the minus one form the power
ok animechic to get rid of a squared term we find the squared root and once you do it too one side of the equ you do to the next side..
ReplyDeleteand another way to wirte square root is a value to the power of ah half
e.g (4x + 3)^0.5 0.5 is the same as 1/2
POINTS TO NOT PPL
ReplyDeletethe heading says find dy/dx right so that means it is with respect to y.
so all u ahve to do is make y the subject of the formula for all the ques in each case and then just Differenate
does anyone still dont understand ????
Question 1.
ReplyDeletey = 6x^3 + 4x^2 - 5x
dy/dx = 18x^2 + 8x - 5
Question 2.
y^2 = x^3 + 16
I need help with this question
Question 5
ReplyDeletey + 6 = 4x + 9y - 3x^2
y = 4x + 9y - 3x^2 - 6
y - 9y = 4x - 3x^2 - 6
-8y = 4x - 3x^2 - 6
y = ( 4x - 3x^2 -6 )/ 6
Now its just a matter of differentiating
so question like these you have to make y subject of the formula and then differentiate normal after.
ReplyDelete1. y = 6x^3 + 4x^2 - 5x
ReplyDeletedy/dx = 18x^2 + 8x - 5
2. y^2 = x^3 + 16
ReplyDeletedy/dx= 3x^2
16 is jus a number dats stands alone hence d differential of it is 0
Question 5.
ReplyDeletey + 6 = 4x + 9y - 3x^2
all y variable grouped &
all x variable grouped
so
y -9y = 4x - 3x^2 - 6
y(-9y) =4x - 3x^2 -6
-9y^2 =4x - 3x^2 -6
y^2 = 4x - 3x^2 - 6/ -9
dy/dx= (4 - 6x)^9
Question 4.
ReplyDeletey = 3x^4 + 3(x^2 + 5)
get rid of the brackets by expanding
so (3*x^2) + (3*5) = 3x^2 + 15
the equation will continue....
y= 3x^4 + 3x^2 + 15
then find dy/dx
dy/dx= 12x^3 + 6x
1.y=6x^3+4x^2-5x
ReplyDeletedy/dx=18x^2+8x-5
2.y^2=x^3+16
dy/dx=2y=3x^2
3.1/y=4x+4
dy/dx=y^-1=4x+4
dy/dx=-1y^-2=4
4.y=3x^4+3(x^2+5) = 3x^4+3x^2+15
dx/dy=12x^3+6x
5.y+6=4x+9y-3x^2
dy/dx=4+9-6x
for 5 ,y + 6 = 4x + 9y - 3x^2 , which is the correct way ,just differentiating or grouping terms..............i think the way weezy did it but not really sure?
ReplyDeletein differentiation, the method is to multiply the coeff. of x by the power of x then minus 1 to the power of x:
ReplyDeletey = ax^n
dy/dx = nax^n-1
1) y = 6x^3 + 4x^2 - 5x
ReplyDeletedy/dx = 3(6x)^3-1 + 2(4x)^2-1 - 5
18x^2 + 8x - 5
Yes "low rider" always make y (or wat is to be differentiated) the subject of the formula before starting to differentiating the equation. Also another thing to take note of is to make put the equation in its simpliest form before differentiating....
ReplyDelete1. 6x^3 + 4x^2 - 5x
ReplyDeletedy/dx =18x^2 + 8x - 5
4. y = 3x^4 + 3(x^2 + 5)
ReplyDeleteRemoving brackets:
y = 3x^4 + 3x^2 + 15
dy/dx = 12x^3 + 6x
5. y + 6 = 4x + 9y - 3x^2
ReplyDeletey = 4x + 9y - 3x^2 - 6
y - 9y = -3x^2 + 4x - 6
-8y = -3x^2 + 4x - 6
Divide both sides by (-8)
y = -3x^2 + 4x - 6 / -8
y = -3x^2 +4x + (3/4)
dy/dx = -6x + 4
1. y = 6x^3 + 4x^2 - 5x
ReplyDeletewhen doing this you multiply the coefficent by the power that value now becomes thr coefficent and 1 is minused from the power the power
y=18X^2+8X-5
2. y^2 = x^3 + 16
ReplyDeletelook for brackets or notted terms
2y=3X^2
when there is a 3 alone the number is removed
5. y + 6 = 4x + 9y - 3x^2
ReplyDeletey (-9y )= 4 - 6X - 6
9Y^2=-2-6X
dy/dx=(18/6)^2
4. y = 3x^4 + 3(x^2 + 5)
ReplyDeletework out the knotted terms and then do normal differentiation
y=12X^3+ 3X^2+15
y=12X^3+6X
dy/dx=12x^3+6x
A differentiation question & answer!!!>>>
ReplyDeleteIf T = (3p + 1)^2/ p ,find dT/dp and the values of p for dT/dp = 0.
Solution^^
Here we rewrite the expression as a polynomial first:
(3p + 1)^2 /p
Dividing each term in the numerators by p
9p^2 + 6p + 1 /p
9p + 6 + 1/p
9p + 6 + p^-1
Then dT/dp = 9 + 0 + (-1)p^-2 = 9 -(1 /p^2)
If dT/dp = 0, then 9 = 1/p^2 and p = +or- 1/3
this 1 iss another eg also
ReplyDeleteDifferentiate wrt x
(2x - 3)^2
Here we express (2x-3)^2 as a polynomial first by expansion::::>so
(2x - 3)^2
4x^2 - 12x + 9
If y 4x^2 - 12x + 9, then
dy/dx = 8x - 12x^0 + 0 =
8x - 12,as x^0=1.
For all of you who tried to ans #2 you should check back your notes that miss give...
ReplyDeleteI think the ans you found were wrong.
y^2 = x^3 + 16
the 1st thing 2 do is to make y the subject of the formula:
y = (x^3 + 16)^0.5
since if you make y the subject then the other side of the equation has to compensate, therefore, in this case the other side becomes the square root or to the power of 0.5
now you diff.
y = (x^3 + 16)^0.5
This is to hard to diff. so im goin to make it simpler as follows:
substitute
m = x^3 + 16
dm/dx = 3x^2
rewrite:
y = m^0.5
dy/dm = 0.5m^-0.5
The question asked for dy/dx.
Therefore:
dy/dx = dm/dx.dy/dm
dy/dx = (3x^2) X 0.5(x^3 + 16)
if im wrong plz correct me...
#5. y + 6 = 4x + 9y - 3x^2
ReplyDeleteMake y the subject.
y = 4x + 9y + 3x^2 - 6
now diff. But remember this is partial diff.
dy/dx = 4 + 9dy/dx - 6x
this is how you do #5...