The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Implicit differentiation help where help stated

linQ - https://twitter.com/natekeegan1986/status/402832415371976704/photo/1

TY -N

Why does the 'x' disappear?
Original post by REASON_Lighters

lna is a number, let's call it c.
ddxcx=c\dfrac{d}{dx}cx=c.
Remember x=x1x=x^{1} and ddxxn=nxn−1\dfrac{d}{dx}{x^n}=nx^{n-1} so ddxx1=1x1−1=1x0=12=1\dfrac{d}{dx}x^1=1x^{1-1}=1x^0=1^2=1.
(edited 10 years ago)
Original post by keromedic
lna is a number, let's call it c.
ddxcx=c\dfrac{d}{dx}cx=c.
Remember x=x1x=x^{1} and ddxxn=nxn−1\dfrac{d}{dx}{x^n}=nx^{n-1} so ddxx1=1x1−1=1x0=12=1\dfrac{d}{dx}x^1=1x^{1-1}=1x^0=1^2=1.


of course differentiating x gets 1 ffs. Thank you extremely clear layout and work, kind regards.

Thanks makes sense in the end :smile:))))))
Original post by REASON_Lighters
of course differentiating x gets 1 ffs. Thank you extremely clear layout and work, kind regards.

Thanks makes sense in the end :smile:))))))

Don't mention it. Quite often it's these things that cost us the marks :h:
Original post by keromedic
Don't mention it. Quite often it's these things that cost us the marks :h:


idd :P

I take it you're doing A2 Maths atm, do you do FM as well?
Original post by REASON_Lighters
idd :P

I take it you're doing A2 Maths atm, do you do FM as well?

I did A2 Mathematics last year. Presently, I'm studying AS/A2 Further Mathematics.
Original post by keromedic
I did A2 Mathematics last year. Presently, I'm studying AS/A2 Further Mathematics.


Very good, me too except I am doing a2 maths this year as well :_)

Sorry to trouble you but I have another question:

R=200(0.9)^t Find the value of dR/dt when t=8

I get dR to be =(200*0.9^t)

But I dont get why you have to multiply by ln 0.9 with my dR :colondollar:
Original post by REASON_Lighters
Very good, me too except I am doing a2 maths this year as well :_)

Sorry to trouble you but I have another question:

R=200(0.9)^t Find the value of dR/dt when t=8

I get dR to be =(200*0.9^t)

But I dont get why you have to multiply by ln 0.9 with my dR :colondollar:

I'm happy to help but do you mind latexing or scanning your question and attempted working?
Original post by keromedic
I'm happy to help but do you mind latexing or scanning your question and attempted working?


yeah sure sorry:


linQ - https://twitter.com/natekeegan1986/status/402845669301305344/photo/1

Well the question is short - and involves product rule, its just that I don't get the 'In 0.9' from the product rule.
Original post by REASON_Lighters
yeah sure sorry:


linQ - https://twitter.com/natekeegan1986/status/402845669301305344/photo/1

Well the question is short - and involves product rule, its just that I don't get the 'In 0.9' from the product rule.

Sorry about the late response. Just checking it now, had to do something.
Original post by keromedic
Sorry about the late response. Just checking it now, had to do something.


No problem, happy for the assistance :smile:
Original post by REASON_Lighters
yeah sure sorry:


linQ - https://twitter.com/natekeegan1986/status/402845669301305344/photo/1

Well the question is short - and involves product rule, its just that I don't get the 'In 0.9' from the product rule.

I'm not sure I understand your question. You've learnt that ddxax=axlna\dfrac{d}{dx}a^x=a^xlna. aa in this case is 0.9. What's the problem?
Original post by keromedic
I'm not sure I understand your question. You've learnt that ddxax=axlna\dfrac{d}{dx}a^x=a^xlna. aa in this case is 0.9. What's the problem?


Thank you, yeah that makes sense, so do you have to multiply by 200 with respect to the product rule as well then?
Original post by REASON_Lighters
Thank you, yeah that makes sense, so do you have to multiply by 200 with respect to the product rule as well then?

200 is a constant so it isn't involved in differentiation. It's not a product rule thing, no.
Original post by REASON_Lighters


Because dydx\frac{dy}{dx} has been omitted, it should be 1ydydx\frac{1}{y}\frac{dy}{dx} om the left hand side.
Original post by brianeverit
Because dydx\frac{dy}{dx} has been omitted, it should be 1ydydx\frac{1}{y}\frac{dy}{dx} om the left hand side.

Ah yes, I hadn't spotted the missing [ted]\dfrac{dy}{dx}. Why/how is that the reason?
(edited 10 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you