Solution 1
y = log base 3 (8x^7 + x)
u = 8 x^7 + x
dy/dx = 56 x^6 + 1
= (56 x^6 + 1) / (8 x^7 + x) log base 3 e
Solution 2
P = ln 6s^3 + s
u = 6s^3
dy/dx = 18 s^2
= (18s^2)/(6s^3) + 1
Solution 2
P = ln 6s^3 + s
u = 6s^3 + s
dy/dx = 18 s^2 + 1
= (18s^2 + 1)/(6s^3 + s)
Solution 3
y = 15 x^3 + 7/(2x^3) - x
y = 15x3 + 7 *2 x^-3 - x
dy/dx = 45 x^2 + 42 x^-4 - 1
Solution 3
y = 15 x^3 + 7/(2x^3) - x
dy/dx = 45 x^2 + 7/6 x^-2 - 1
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82 comments:
i think in solution 1 there is an error. if you were to carefully work this question according to this formula; dy/dx = 1/u(du/dx)log[3] e
the error in this solution is in the last line. the final answer supposed to be:
dy/dx = (8x^7 + x)/(56x^6 + 1) log[3] e
solution 2 is basically correct meaning that dy/dx needs to be replaced with ds/dp because that is the terms given in the equation.
and the +1 needs to remain +s because there is only one thing being differentiated because there is no brackets present.
Tweety, i believe that you are incorrect.if memory serves me correctly. du/dx and the part that is defrentiated is always on top as you say 'according to the formula'.Take a reallly good look at it. I think question one is correct. However you should write the formula you are useing in your statement, when you are solving a question, you shouldn't just assume that the person who is correcting knows what approach to solving you are useing.
There are a couple doubts about the first two questions that ar bugging me though...In question 1 should the 1 be there in the final answer? And the final line in question 2 isn't simplificaqtion required here or something??
If you solve 2 is the final answer 18+s/6+s^2?
the first one is incorrect. after finding u. dy/ddx is not the next step. the nex step is du/dx
u
will solution 1 is correct and tweety u made a critical error with the formula so i would agree with crimson crock on that defrentiated is always on top.
n2o its actually dP/ds not ds/dP since you are differentiating for P with respect to s.
No moose heres the correct solution for the first ques.3
y = 15x^3 + 7/(2x^3) - x
y = 15x^3 + 7/2 x^-3 - x
dy/dx = 45x^2 - 10.5x^-4 - 1
They all have errors!!!
copy cat i also think solution 1 is incorrect since the differentiation should of been du/dx not dy/dx because that is what we are trying to find
for solution 2.
P=ln6s^3+s
u=6s^3+s
du/ds=18s^2+1
dy/dx=1/u *du/ds
dy/dx=18s^2+1/6s^3+s
Crimson Crock in resonse to your 3rd comment when you solve ques.2 you should get-:
dP/ds = (18s^2)/(6s^3) + 1
i still have not really understood ln when it comes to differentation
and n2o whilst i appreciate your comment i'm still kinda lost. Can anyone explain?
only after posting my previous comment i noticed tinkerbell's comment, in following the steps i think it kinda cleared things up
Wat u all tink about solution 3. I tink it is wrong.
y = 15 x^3 + 7/(2x^3) - x
dy/dx = 45x^2 + -10.5x^-4 - 1
And u want 2 kno how i get -10.5x^4rite??? its very simple. If u are given 7/2x^3. This can be re-expressed as 7/2x^-3 since the x^3 in the denominator will go to the numerator. U can then differentiate normally..
tinkerbell you diffrentiated the +s to be +1 and then came back and put +s!!! why???
For the second sol'n 3 isn't it that you have to use the
quotient
rule?
In solution 3.
y=15x^3+7/(2x^3)-x
Y=15x^3+(7)(2x^-3)-x
y=15x^3+14x^-3-x
dy/dx=45x^2-42x^-4-x
tinkerbell you need to review wat u doin you making some dred errors!!!!
i think that solution 2 is wrong in the last line forgot (loge) and i dont think that the +1 is not have to be there, also in dy/dx is wrong because u forget to include (s).
in solution (2a) it is quite different where the second line is wrong because (s) is not loged so u cannot add it in (u)and the dy/dx is correct.
Well lilo responing to your 2nd comment when ln comes into the eqn. you simply diffrentiate has you would with the log but you leave out log e for example-:
y = log 2x^2
dy/dx = 4x/2x^2 log e
y = ln 2x^2
dy/dx = 4x/2x^2 ...dats as easy as it gets.
i agree with crimson crock. the equation is actually dy/dx=(1/u)du/dx loge. therefore the answer would result to be
du/dx= 56x^6 + 1
dy/dx =(1/(8x^7 + x))(56x^6 + 1)x log[3]e
or=(56 x^6 + 1) / (8 x^7 + x) log base 3 e
TINKERBELL look here!!!
7/2x^3 = 7*2 x^-3 (WRONG)
7/2x^3 = 7/2 x^-3 (CORRECT)
i believe solution 2 is incorrect because the equation was p in terms of s and since there was no dp/ds, there would not be any differential.
the first one is definetely incorrect after looking at that example since according to what miss taught me last week it should be du/dx
In response to crimson crock:
in questions 1, the 1 is found by differentiating x.
the final line in questinon two can be simplified :
(18s^2)/(6s^3)=[6s^2(3)]/[6s^2((s)]
cancel the 6s^2 and you get
3/s
with respect to solution 1. i tink its correct. Most of u who commented on solution 1 said that it is wrong why. Ent the formula for dy/dx is 1/u (du/dx)log[3] e. wait! wait! wait! is du/dx not dy/dx. solution 1 is wrong because of the line suppose to be du/dx. The calculations are correct though!!!! Wat u all tink???
also the dy/dx is incorrect because it should be
du/dx=56 x^6 + 1
and the answer must be represented in dy/dx=(1/u)(du/dx)loge
my last comment applies to question 1
y = log base 3 (8x^7 + x)
in number 1 the third line is wrong.... instead it is
du/dx=56 x^6 + 1
therefore usin the formula.... dy/dx=1/u * du/dx log e
which is equal to(56x^6 +1)/8x^7 + 1 log base 3 e
For solution 1
y = log base 3 (8x^7 + x)
u = 8x^7 + x
and most importantly it is du/dx not dy/dx if you do this miss is going to give you ZERO
Therefore
du/dx = 56x^6 + 1
and dy/dx = (56x^6 + 1)/(8x^7 + x)log base 3 e
this is the same mistake for solution 1 and both solution 2, where dy/dx was placed instead of du/dx.
eh bakes see wat going wit this question na all yuh
3/4 x^8 - 1/2
y=3(4x^-8) - 1/2
dy/dx= 3(-32x^7)
dy/dx= -3/32x^7
in solution 1 du /dx=56x^6+1, not dy/dx because it was differentiation with respect to u.
Both questions on solution 2 is wrong. for the 1st question, again the third line suppose to be du/dx and not dy/dx. For the second question on solution 2 u = to 6s^3 and not 6s^3 + s. since there are no brackets. Again the third line suppose to be du/dx and not dy/dx.
In Solution 1 all the working is correct but an error lies in the syntax in line 3. The correct syntax should be du\dx not dy\dx
i have a question.
If P=ln 7s^3 + 3s^2 Differentiate P with respect to s
for the 1st solution 2, when i read the question i couldn't find the error but after going through it in class today i realised the error lies in the third line, where the differential was dy/dx instead of dp/ds where the instantaneous rate of change of pressure p was to be found
y = 3/4x^8 - 1/2
dy/dx = 6x^7
or dy/dx = 6x^-9
i'm not qutie sure because it is not clearly seen as to where the x is placed correct me if i'm wrong
solution 1 is incorrect because dy/dx is not 56x^6 + 1.du/dx is 56x^6 + 1. the final answer is correct but the solution is wrong
P = 7s^3 + 3s^2
dP/ds = 21s^2 + 6s
is that the correct answer cokebaby
for solution 1 the third line is wrong, the differential should be du/dx instead of dy/dx since the function is based on u and not y.
where the problems have an x by itself at the end why would you say it can be understood as 1??
Oh gosh so allyuh go eat mih head
I now see my mistakes and i stand corrected.
For solution 2
P=ln6s^3=3
u=6s^3
du/dx=18s^2
dP/ds=18s^2/6s^3+1
Solution 3
y=15x^3+7/2x^3-x
dy/dx=45x^2-21/2x^-4-1
Here is forty five x squared minus twenty-one over two x to the power of minus four,minus one.
So you have to pay to learn
Ilook stupid.only for a while but I learn!!!!!
hey ppl wats up i think the mathematical part of this answer in questionis correct but wat are u trying to find difinitely not dy/dx wat u are tryin to find is the rate of change of distance with respect to pressure correct so therefore it shud be ds/dp = (18s^2)/(6s^3 + s)
So all the solutions were incorrect. Go easy on me nah Mr. Psychotic!!!
thanks alot psychotic it cleared things up in ln for me
aye moose, the question is
y=3/(4x^8) - 1/2, right? well i think that it can be broken down to( 3/4 x 1/x^8 )-1/2
dy/dx= -8(3/4)x^-9
dy/dx= -24/4x^9
dy/dx= -6x^9
hEY YOU FOLKS i JUST WORKED OUT THE FIRST ONE i AM VERY CONFUSED AFTER READING SOME OF YOUR COMMENTS. If I my calculations serves me right The first one is correct
Today in class I thnk i finally grasped the concept of finding the derivitive of these questions especiallly where we have the fractions
shouldn't the differential in solutions 2 be expressed as dP/ds?...im not sure..im just thinking out loud here
the answer for solution 3 is incorrect, the correct answer would hav been dy/dx=45x^2-(42x^-4)-1 or dy/dx=45x^2-(7/6x^4)-1. but the reason i think u got it wrong in the first place is bcause u didnt go each step by step whereby changing 7/2x^3 to 7(2x^-3), and then differentiate in order so see i u are making a mistake
I have a problem!!! i have a project to do for miss. can u all give me sum real life applications of Logarithms and complex numbers in Chemical Engineering?????
I have a question agiain!! Wat is the significance of the y-intercept in real life problems?
In relation to complex numbers, wat is the imaginary part and waas is the real part? Im comfused!!!!!!
solution 1
this is wrong because when you differentiate u you suppose to have du/dx and not dy/dx
Solution one seems to be qutie correct. It follows the basic laws involving the diffrential of logs where differiate whatsoever logs there is in backets then whatsoever value you get you divide over whatsoever value you had to diffrentiate.
solution 1 and solution 2 has the same error where they have dy/dx. to the untrained eyed it may look correct but if you look closely u would c the question has u in it therefor it should b for solution1 du/dx an for solution 2 it should b du/ds.
In solution 1 there is just a simple error in which instead of saying du/dx they said dy/dx
soluton 2
this is also wrong because u suppose to be 6s^3 + s. therefore you will get a completely different answer also you supposed to be du/dx and not dy/dx
I do not think that there is nothing wrong with how question one is done. It seems that the solution to this question was done using the basic law of diffrentiation when it comes to logs where the diffrential of the expression in brackets is placed over the expression itself
solution 3
this is completely wrong. because you can't have 7/(2x^3) and multiply them.
Solution 1 dy/dx=56x^6+1 is not correct it should be du/dx=56x^6+1
The same thing with solution 2 dy/dx=18s^2 it should be
du/ds=18s^2
solution one looks correct
sol 2 u is incorrect also it is dp/ds
sol 3 dy/dx =(45x ^2 - ^-4 -1)/
(15x ^3 +7/2x ^3 -x)og e
All of the solutions 1,2,3 are wrong
in solution 1..it is not suppose to be dy/dx, it is suppose to be in respect to the equation..therefore it is to be du/dx...
NO: 2 is not correct meaning that dy/dx needs to be replaced with ds/dp because that is the terms given in the equation.
solution 1,2, and 3 are wrong from the third line down because it is du/dx, not dy/dx.from that little mistake, the answer is inaccurate.
I agree that it is dP/ds not ds/dP since you are differentiating for P with respect to s.concerning solution's 2.
concerning question 3:
y=15x^3+7/(2x^3)-x
Y=15x^3+(7)(2x^-3)-x
y=15x^3+14x^-3-x
dy/dx=45x^2-42x^-4-1
yea the master here again. coming to these solutions, i will go through everyone step by step and also point out the errors that was incorporated into the questions.
question 1:
y = log base 3 (8x^7 + x)
let u be equal to 8x^7 + x
u = 8 x^7 + x
du/dx = 56x^6 + 1
at this point if u look back at the previous answer u will of seen:
dy/dx = 56x^6 + 1..... THIS IS TOTALLY WRONG. U MIGHT SAY WHY IT IS WRONG, THEY BOTH HAVE THE SAME ANSWER...... but if u look carefully you will see that 'u' is being differentiate with respect to 'x' and the differential of 'u' is du/dx and NOT dy/dx. well going on:
dy/dx = 1/u du/dx log e
dy/dx=(56x^6 + 1)/(8 x^7+ x)log[3]e
question 2:
P = ln 6s^3 + s
let:
u = 6s^3
du/ds = 18s^2
remember we are differentiating
'u' with respect to 's' so du/dx or dy/dx IS VERY WRONG.. IT DOESN'T HAVE ANY VARIABLE 'Y' OR 'X' IN THIS EQUATION.anyhow...
dP/ds = 1/u du/ds (1) + 1
dP/ds = (18s^2)/(6s^3) + 1
dP/ds = 3/s + 1
question 3:
y = 15 x^3 + 7/(2x^3) - x
y = 15 x^3 + (7/2)x^(-3) - x
dy/dx = 45x^2 - (21/2)x^(-4) - 1
dy/dx = 45x^2 - 21/2x^4 - 1
hope these little tips helps anyone out there. laterz....
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