The Student Room Group

Scroll to see replies

u = sin(x)
Reply 2
DeanK22
u = sin(x)


Hmmm, I'm sorry I don't know what to do with that..
Rei
Hmmm, I'm sorry I don't know what to do with that..


oh. substitution.

I thought it would make it a bit more obvious as to what to do

Really all you are showing using that sub is that;

Spoiler

Reply 4
Try the quotient rule if you're not used to substitution.
Differentiating? First times the differential of the second, subtracted from the second times the differential of the first, all over the second squared.

The Quotient rule.
JMonkey
Try the quotient rule if you're not used to substitution.


shamrock92
First times the differential of the second, subtracted from the second times the differential of the first, all over the second squared.

The Quotient rule.


feel free to tell me how this is relevant at all to the problem in hand?
Reply 7
DeanK22
feel free to tell me how this is relevant at all to the problem in hand?


Do you want me to solve it and then send you a pm. I just thought if he's not got to substitution yet then it's soluble with the quotient rule and it is. Believe me, I usually solve it on paper before I post or at least look at the solution on a maths program on my PC if its obvious and the answer gels with that solution type.

It is relevant because it is a method of solution if you don't know certain rules yet.

I'm well aware of your method being easier. That's why I posted the quotient rule, because he didn't get u=.
DeanK22
feel free to tell me how this is relevant at all to the problem in hand?


Presumably, what the OP means is: "How do I differentiate 40cosx53sinx\frac {40cosx}{5-3sinx}?" Otherwise he's just made a statement.

f(x)=g(x)h(x)    f(x)=g(x)h(x)h(x)g(x)[h(x)]2f(x) = \frac{g(x)}{h(x)} \implies f'(x) = \frac{g'(x)h(x) - h'(x)g(x)}{[h(x)]^2}

So apply that rule.
JMonkey
Do you want me to solve it and then send you a pm. I just thought if he's not got to substitution yet then it's soluble with the quotient rule and it is. Believe me, I usually solve it on paper before I post or at least look at the solution on a maths program.

It is relevant because it is a method of solution if you don't know certain rules yet.


it was a rhetorical question. It is infact useless advice and has possible confused the person who sought help.
shamrock92
Presumably, what the OP means is: "How do I differentiate 40cosx53sinx\frac {40cosx}{5-3sinx}?" Otherwise he's just made a statement.

f(x)=g(x)h(x)    f(x)=g(x)h(x)h(x)g(x)[h(x)]2f(x) = \frac{g(x)}{h(x)} \implies f'(x) = \frac{g'(x)h(x) - h'(x)g(x)}{[h(x)]^2}

So apply that rule.


dydx=f(x) \frac{dy}{dx} = f(x) is an equation and i gathered from my knowledge of further maths module 1 that y was reuired as a function of x.

If he wanted to differentiate he would have stated;

ddx40cosx53sinx=f(x) \frac{d}{dx} \frac{40cosx}{5-3sinx} = f(x) what is f(x) ? .
Reply 11
DeanK22
it was a rhetorical question. It is infact useless advice and has possible confused the person who sought help.


I'd hold off second guessing the OP until he replies.

It might be that he tried it that way and got stuck. He asked how to start it off. We gave him the answer to what he asked. Set it up in quotient rule form. And go from there.
JMonkey
I'd hold off second guessing the OP until he replies.

It might be that he tried it that way and got stuck. He asked how to start it off. We gave him the answer to what he asked. Set it up in quotient rule form. And go from there.


sigh.

this is so very tiresome. Do you know that differentiation and integration are different processes and how your advice in this thread is so far anything but ?
Reply 13
DeanK22
dydx=f(x) \frac{dy}{dx} = f(x) is an equation and i gathered from my knowledge of further maths module 1 that y was reuired as a function of x.

If he wanted to differentiate he would have stated;

ddx40cosx53sinx=f(x) \frac{d}{dx} \frac{40cosx}{5-3sinx} = f(x) what is f(x) ? .


That's Liebniz notation for differentiate though? Not sure what you are getting at? Alot of my texts use that shorthand for differentiate with respect to?

Derivative dy/dx.
JMonkey
That's Liebniz notation for differentiate though? Not sure what you are getting at? Alot of my texts use that shorthand for differentiate with respect to?


I am not sure what you are getting at - how my choice of notation is worrying you is confusing, let alone the fact that you are focusing on my choice of notation and not the actual sentences before;

theop wanted to integrate.

IF he wanted to differentiate; (d/dx) f(x) = g(x) find g(x)
Reply 15
DeanK22
sigh.

this is so very tiresome. Do you know that differentiation and integration are different processes and how your advice in this thread is so far anything but ?


Stop patronising me thank you. That is the terminology for a differential if he wanted to integrate it he's using the wrong terminology.

Honestly? What?

Derrivative of y with respect to x= dy/dx?
JMonkey
Derivative dy/dx.


what?
Reply 17
DeanK22
I am not sure what you are getting at - how my choice of notation is worrying you is confusing, let alone the fact that you are focusing on my choice of notation and not the actual sentences before;

theop wanted to integrate.

IF he wanted to differentiate; (d/dx) f(x) = g(x) find g(x)


He said differential question in the title, you have lost me now?

I'm willing to concede that you have the text in front of you it is an integral, but its hardly clear from the op.
JMonkey
Stop patronising me thank you.


I think it should be noted that I started condescending you (that is different from patronising) AFTER you blatently ignored my correct description of this problem.

I am in no way sounding arrogant here but I can firmly say my knowledge of this topic is beyond yours and you are merely wasting my time and distracting me when you post incorrect and downright nonsensical posts in a thread that i answered.
From the first post of the thread it seemed like the original poster wanted to solve a certain differential equation, in which case the substitution in the second post would have been relevant. I guess this is what happens when people don't post the entire question...

Latest