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Question 1
ReplyDeleteThe equation of a curve is y = 2x + 8/x^2
find dy/dx and d^2ydx^2
y = 2x + 8/x^2
First thing to do is to make the equation simplier
y = 2x + 8x^-2
dy/dx = 2 - 16x^-3
d^2y/dx^2 = the differntiation of dy/dx
= 48x^-4
Question 2
ReplyDeleteDifferentiate ln (2x + 3)
The rule states that when
y = ln x...................dy/dx= 1/x
y = ln (2x + 3)
Make a subsitution
let m = 2x + 3
dm/dx = 2
Rewrite
y = ln m
dy/dm = 1/m
Therefore dy/dm = dm/dx x dy/dx
= 1/m x 2
but m = 2x + 3
So
dy/dx = 1/(2x + 3) * 2
Question 3
ReplyDeleteA curve is such that dy/dx = 2x^2 -5. Given that the point (3,8) lies on the curve, find the equation of the curve.
First we find the intergral of dy/dx
y = 2x^3/3 - 5 + C
Now substite the points in the equation (3,8)
where y = 8................x = 3
8 = 2(3)^2/3 - 5 + C
Someone correct me if am on the wrong stract plez...
equation of a curve is an exponential
ReplyDeleteif dy/dx = 2x ^ 2 - 5 ( this is the tool for the gradient)
thus the gradient at point (3,8) is
= 2 (3^2) - 5
= 13
the integral of dy/dx = 2x^2 - 5x is
= 2x^3/3 - 5x + c
thus y = 2x^3/3 - 5x + c
ok i think i sooooo lost now .....!!!
Differentiate ln (2x + 3)
ReplyDeleteThe rule states that when
y = ln ( x )dy/dx= 1/x
y = ln (2x + 3)
Make a subsitution
let m = 2x + 3
dm/dx = 2
Rewrite
y = ln m
dy/dm = 1/m
Therefore dy/dm = dm/dx x dy/dx
= 1/m x 2
but m = 2x + 3
So
dy/dx = 1/(2x + 3) * 2
Miss Fariel i forgot to give my name to you . this is my I.D. number 110000239
ReplyDeleteok swamky thing when u integrate you are doing it wrong remember the term 5 can also be integrated and it will become 5x
ReplyDeleteso when integrated you will get
y = 2/3 x^3 - 5x+ C
and i beleive that is the answer
now u can sub the pts (3,8) in to the eqa
and johnsmith the answer u got for C is wrong so check it over
8 = 2/3(3^3) - 15 +C
8 = 18 - 15 + C
8 = 3 + C
c = 5
now rewirte eqa
y = 2/3 x^3 - 5x + 5
y = 2x + 8/x^2
ReplyDeleteNo downstairs
y = 2x + 8x ^-2
y = 2x + 16x
Differentiate ln (2x + 3)
ReplyDeleteRewrite ln m
dy/dm = 1/m
Let m = 2x + 3
dy/dx = 1/ 2x + 3
cnckprpl am trying to understand what u did for your first que and its quite confusing .....
ReplyDeletethe first time you find dy/dx you should get
y = x + 16x^-3
when u minus one from minus 2 you will get minus 3
cnckprpl your second question is also wrong
ReplyDeletewhen u Differentiate 2x + 3 you will get 2
soo it will be dm/dx X dy/dm
so your answer will be 1/2x+3 x 2
ln (2x+3)
ReplyDeleteyou look to see which rule you are dealing with.
and ln x = dy/dx= 1/x
so dy/dx = 1/2
Question 1
ReplyDeleteThe equation of a curve is y = 2x + 8/x^2
find dy/dx and d^2ydx^2
firstly bring the downstairs upstairs
so y = 2x + 8x ^-2
so dy/dx = 2 - 16x
d^2ydx^2, that u just need to Differentiate a second time..
so d^2ydx^2= -16
Differentiate ln (2x + 3)
ReplyDeletewhen y = ln ( x )dy/dx= 1/x
y = ln (2x + 3)
let m = 2x + 3
dm/dx = 2
y = ln m
dy/dm = 1/m
m = 2x + 3
So
dy/dx = 1/(2x + 3) * 2
ln(2x+3)
ReplyDeletey = ln ( x )dy/dx= 1/x
y = ln (2x + 3)
Make a subsitution
let m = 2x + 3
dm/dx = 2
Rewrite
y = ln m
dy/dm = 1/m
Therefore dy/dm = dm/dx x dy/dx
= 1/m x 2
but m = 2x + 3
So
dy/dx = 1/(2x + 3) * 2
y = 2x + 8/x^2
ReplyDeletey = 2x + 8x ^-2
y = 2x + 16x
Ok "renijen" i get where your coming from with question 1
ReplyDeleteln (2x+3)
ReplyDeleteRecall lnx=1/x
hence
ln(2x+3)= 1/(2x+3)x d/dx(2x+3)
= 2/(2x+3)
i'll try 1 as well
ReplyDelete1) y = 2x + 8/x^2
dy/dx=2 -16x^-3
and d^2ydx^2= 84x^-4
1. The equation of a curve is y = 2x + 8/x^2
ReplyDeletefind dy/dx and d^2ydx^2
solution:
y = 2x + 8/x^2
dy/dx = 2 + (-16x^-3)
d^2y/dx^2 = -48x^-4
Differentiate ln (2x + 3)
ReplyDeletesolution:
differential of lnx = 1/x
let (2x + 3)= n
dy/dn = 2
dy/dx = 1/2x+3*2
1. The equation of a curve is y = 2x + 8/x^2
ReplyDeletefind dy/dx and d^2ydx^2
this just means that after you find dy/dx you have to differentiate again.
NB: remember to bring all downstairs upstairs.
y = 2x + 8x^-2
dy/dx = 2 - 16x^-3
then again:
d^2(y)/dx^2 = 48x^-4
Ok guys... Maybe this might help you understand #2.
ReplyDeleteNB: integration > power is increased by 1 and then the coefficient is divided by the power.
dy/dx = 2x^2 - 5
when integrated: y = 2x^3/3 - 5x + c
Now the question asked for the equation of the curve in which the point (3,8) lies on the curve.
so we substitute x=3 and y=8 into the equation we just integrated to find c (y-intercept).
8 = 2(3)^3/3 - 5(3) + c
8 = 18 - 18 + c
c = 8
now rewrite: y = 2x^3/3 - 5x + 8 => equation of the curve.