Derivatives

[Khoxy]

I need a clue on how I can differentiate the following

X^3 + y^2 - 5xy = 0

[Dubbelosix]

I need a clue on how I can differentiate the following

X^3 + y^2 - 5xy = 0

 

Ok, so what do you want to differentiate with respect to?

 

Let's say you want to differentiate with respect to time for instance?

[Dubbelosix]

You're not very quick to reply, so I'll give a demonstration. Rearrange by adding the minus quantity:

 

[math]x^3 + y^2 = 5xy[/math]

 

and say you you know the differential

 

[math]\frac{d}{dt}[/math]

 

You distribute the differential through

 

[math]\dot{x}x^2 + \dot{y}y = 5\frac{d(xy)}{dt}[/math]

 

The dot over the variables denotes you are taking

 

[math]\dot{x} = \frac{dx}{dt}[/math]

 

As a simple demonstration.

Edited June 20, 2019 by Dubbelosix

[Dubbelosix]

Say now you want to take twice the differential, this allows you to write it in the following way

 

[math]\frac{d}{dt}[/math]

 

You distribute the differential through

 

[math]\dot{x}^2x + \dot{y}^2 = 5\frac{d^2(xy)}{dt^2}[/math]

Edited June 20, 2019 by Dubbelosix

[Khoxy]

Thanks Dubbelosix. I have taken time to respond due to slow mobile network in my country.

[GAHD]

https://www.khanacademy.org/math/ap-calculus-ab/ab-differential-equations-new/ab-7-1/v/differential-equation-introduction

If you can get past this guy's repeating the same word several times while writing, this is a good place to get some help understanding it. It would probably help you to understand what a differential actually is so you can reason though why you want to use them and figure out what you're actually doing by plugging things in. Figure out what you're doing, and the how will come a little more natural. It all comes down to ratios.
https://www.khanacademy.org/math/ap-calculus-ab/ab-differential-equations-new

[Dubbelosix]

I'll try my best to teach as well... hopefully I can be clear enough on how it is done. 

[Vmedvil2]

You're not very quick to reply, so I'll give a demonstration. rearange by adding the minus quantity:

 

[math]x^3 + y^2 = 5xy[/math]

 

and say you you know the differential

 

[math]\frac{d}{dt}[/math]

 

You distribute the differential through

 

[math]\dot{x}x^2 + \dot{y}y = 5\frac{d(xy)}{dt}[/math]

 

The dot over the variables denotes you are taking

 

[math]\dot{x} = \frac{dx}{dt}[/math]

 

As a simple demonstration.

 

No offense dubbel but shouldn't there be a 3 in front of that X using the exponent rule for derivation and a 2 in front of the Y being that Y² and X³ was the original variable. I think you forgot the exponent rule.

Edited June 20, 2019 by VictorMedvil

[Dubbelosix]

Well yes, you are right...

 

 

for instance, the differential of [math]x^3[/math] means you lower the power by one and place the exponential as the constant so we would have

 

[math]3 \dot{x}x[/math]

[Dubbelosix]

Now... let's concentrate on the differential of the variable x. I'll write the tex out now.

[Vmedvil2]

Well yes, you are right...

 

 

for instance, the differential of [math]x^3[/math] means you lower the power by one and place the exponential as the constant so we would have

 

[math]3 \dot{x}x[/math]

 

Thats what I thought, but I was never very good at calculus, Exponent Rule.

 

 

This is why I like to use summations instead of derivatives.

Edited June 20, 2019 by VictorMedvil

[Dubbelosix]

[math]\frac{d}{dx}(x^3 + y^2 + 5xy) = 3x^2 + 5xy[/math]

so let's see how you do it - first you apply the differential rule

[math](a \pm b)' = a' \pm b'[/math]

which is the generic formula, so we apply this to your equation

[math]\frac{d}{dx}(x^3) + \frac{d}{dt}(y^2) + \frac{d}{dx}(5xy)[/math]

[math]\frac{d}{dx}(x^3) = 3x^2[/math]

as victor rightly pointed out.

the differential with respect to another variable is always zero

[math]\frac{d}{dx}(y^2) = 0[/math]

But the differential to

[math]\frac{d}{dx}(5xy) = 5y[/math]

so why is this? The steps to understand why goes like this:

[math]= 5y \frac{d}{dx}(x)[/math]

and since

[math]\frac{d}{dx}(x) = 1[/math]

is a function of itself, then we have

[math]=5y \cdot 1[/math]

Plugging in all the terms we have

[math]=3x^2 + 0 + 5y = 3x^2 + 5y[/math]

[Dubbelosix]

Thats what I thought, but I was never very good at calculus.

 

 

 

Don't go onto the ''or'' part because that is a bit too complicated for the other people to understand at this point, who needs a simple demonstration.

[Vmedvil2]

Don't go onto the ''or'' part because that is a bit too complicated for the other people to understand at this point, who needs a simple demonstration.

 

That is fine, I will put it like this or the fundamental theorem of calculus can be used which is.

 

[Dubbelosix]

That's all very good, but to someone who is not very good at calculus, that equation can look very daunting.

[Vmedvil2]

That's all very good, but to someone who is not very good at calculus, that equation can look very daunting.

 

Agreed I remember having to solve it test 1 of calculus 1 and show the proof of the exponent rule which was a pain in the butt.

Edited June 20, 2019 by VictorMedvil

[Dubbelosix]

I forgot that I was writing a tutorial for examples in calculus that relates to physics, you can follow the update here and previous examples which you may find interesting:

 

https://calculusalgebra.quora.com/