Help in Calculus

[Silent Fang]

Hey everyone im really struggling in calculus we are doing derivatives at the moment and ive missed some days due to a close friend dieing but anyhow im not really sure what the product rule is, chain rule, and implicit differential are. Any help(preferably examples-i find i learn best from seeing how others do it) would be greatly appreciated.

 

[Koot]

Do you know how to find derivatives using the definition? The definition is the limit, this should be learned as the most basic way of finding derivatives. The rules are just shortcuts.

Product rule is

y = uv
y' = u'v + uv'

Say y = xlnx

y' = x 1/x + (1)lnx
y' = lnx + 1


Chain rule is used for powers
y = u^c
y' = cu(u')^c-1

Say y = (2x + 1)^4

y' = 4(2x + 1)^3 * 2 <--- 2 is from deriv of 2x+1


Implicit is a bit more advanced as you should know the above first. It's taking the derivative when there are 2 variables on either or both sides. Like with functions that are circles and such.

y = x^2 + y^2
y' = 2x + 2yy'

If you take the derivative of x, it is 1. But when you take the derivative during implicit of y it becomes y'. So you do that on the side that also has x variables. Then you can combine everything and get y' all on one side to get the equation for the derivative.

You should also need the quotient rule if you haven't.
y = u/v
y' = u'v - uv' / v^2

y = (x+2) / x^2
y' = (1)(x^2) - 2x(x+2)
..............x^4

 

[confuego116]

Do you know how to find derivatives using the definition? The definition is the limit, this should be learned as the most basic way of finding derivatives. The rules are just shortcuts.

Product rule is

y = uv
y' = u'v + uv'

Say y = xlnx

y' = x 1/x + (1)lnx
y' = lnx + 1


Chain rule is used for powers
y = u^c
y' = cu(u')^c-1

Say y = (2x + 1)^4

y' = 4(2x + 1)^3 * 2 <--- 2 is from deriv of 2x+1


Implicit is a bit more advanced as you should know the above first. It's taking the derivative when there are 2 variables on either or both sides. Like with functions that are circles and such.

y = x^2 + y^2
y' = 2x + 2yy'

If you take the derivative of x, it is 1. But when you take the derivative during implicit of y it becomes y'. So you do that on the side that also has x variables. Then you can combine everything and get y' all on one side to get the equation for the derivative.

You should also need the quotient rule if you haven't.
y = u/v
y' = u'v - uv' / v^2

y = (x+2) / x^2
y' = (1)(x^2) - 2x(x+2)
..............x^4

these are correct,but there are others you need to know...
Also:

Exponential Rule:
Example: D/DX[e^x] = e^x * lne * 1--the one being the derivative of x(using power rule)
In other words... its becomes itself, which usually doesn't happen except for a few particular cases.

Log rule:

example: D/DX[lnx] = (1/x) * (1/lne) * (1)--the one being the derivative of x(using power rule)
thus: y' = 1/x;


D/DX[sinx]=cosx;

D/DX[cosx]=-sinx;

D/DX[tanx]=sec^2x;

D/DX[cotx]=-csc^2x

D/DX[secx]=secxtanx

D/DX[cotx]=-cscxcotx

also, when doing chain rule with trig f[x]'s this may help as an example:

D/DX[sin^6x]= (sinx)^6;

y'=6(sinx)^5 * cosx or... 6sin^5x * cosx

#note: the angle NEVER changes(unless your doing double angle theorem)

Also, if you're taking AP calc(i'm assuming you are) chain rule may be applied to the product rule which would require finding GCFs after finding the rate of change.


if you need any help ask

 

[Silent Fang]

these are correct,but there are others you need to know...
Also:

Log rule:

example: D/DX[lnx] = (1/x) * (1/lne) * (1)--the one being the derivative of x(using power rule)
thus: y' = 1/x;


D/DX[sinx]=cosx;

D/DX[cosx]=-sinx;

D/DX[tanx]=sec^2x;

D/DX[cotx]=-csc^2x

D/DX[secx]=secxtanx

D/DX[cotx]=-cscxcotx

also, when doing chain rule with trig f[x]'s this may help as an example:

D/DX[sin^6x]= (sinx)^6;

y'=6(sinx)^5 * cosx or... 6sin^5x * cosx

#note: the angle NEVER changes(unless your doing double angle theorem)

Also, if you're taking AP calc(i'm assuming you are) chain rule may be applied to the product rule which would require finding GCFs after finding the rate of change.


if you need any help ask

thanks for both of the answers and i completely forgot we did the trig ones too and thanks because i didnt have those in my notes. Also i do know how to find the derivative the long way we spent awhile on it and now we can just use the shortcut method, ill look through what you guys posted, thanks again for the help

 

[Silent Fang]

Do you know how to find derivatives using the definition? The definition is the limit, this should be learned as the most basic way of finding derivatives. The rules are just shortcuts.

Product rule is

y = uv
y' = u'v + uv'

Say y = xlnx

y' = x 1/x + (1)lnx
y' = lnx + 1


Chain rule is used for powers
y = u^c
y' = cu(u')^c-1

Say y = (2x + 1)^4

y' = 4(2x + 1)^3 * 2 <--- 2 is from deriv of 2x+1


Implicit is a bit more advanced as you should know the above first. It's taking the derivative when there are 2 variables on either or both sides. Like with functions that are circles and such.

y = x^2 + y^2
y' = 2x + 2yy'

If you take the derivative of x, it is 1. But when you take the derivative during implicit of y it becomes y'. So you do that on the side that also has x variables. Then you can combine everything and get y' all on one side to get the equation for the derivative.

You should also need the quotient rule if you haven't.
y = u/v
y' = u'v - uv' / v^2

y = (x+2) / x^2
y' = (1)(x^2) - 2x(x+2)
..............x^4

Koot I want to thank you for these examples i think ive got the chain rule because I was in that day but my teacher made the other two seem so confusing(from what my friends said) im gona try a few problems from the book and see how it goes, thanks again

edit:i somehow think i learned the quotient rule one of those days and my teacher always says, low d-high minus high d-low all over low^2
edit2: alright actually im off to bed ill look at some of the problems out of the book tomorrow, thanks again guys