# Confusing Differentiation

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How do you differentiate (x/v) + (√(a-x)^2+(a/2)^2)/(v/3) WRT x where v and a are constants?

Edit: Don't worry I've solved it.

Edit: Don't worry I've solved it.

Last edited by domm1; 4 weeks ago

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#2

(Original post by

How do you differentiate (x/v) + (√(a-x)^2+(a/2)^2)/(v/3) WRT x where v and a are constants?

**domm1**)How do you differentiate (x/v) + (√(a-x)^2+(a/2)^2)/(v/3) WRT x where v and a are constants?

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(Original post by

First term is easy, chain rule for the second. Have a go?

**mqb2766**)First term is easy, chain rule for the second. Have a go?

P.S. I have not learnt the Chain Rule yet.

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#4

(Original post by

Yeah I was being stupid as I was trying to get an expression for x in terms of a and v when the derivative was equal to 0, but ended up with an equation that was of the form of a quadratic but in terms of x, a and v, but already have the values for a and v, oops.

P.S. I have not learnt the Chain Rule yet.

**domm1**)Yeah I was being stupid as I was trying to get an expression for x in terms of a and v when the derivative was equal to 0, but ended up with an equation that was of the form of a quadratic but in terms of x, a and v, but already have the values for a and v, oops.

P.S. I have not learnt the Chain Rule yet.

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(Original post by

Do you want to describe what the problem is / upload your working?

**mqb2766**)Do you want to describe what the problem is / upload your working?

Part B. I am just worried I have differentiated the second term incorrectly.

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#6

(Original post by

https://isaacphysics.org/questions/l...a-1ea18244028f

Part B. I am just worried I have differentiated the second term incorrectly.

**domm1**)https://isaacphysics.org/questions/l...a-1ea18244028f

Part B. I am just worried I have differentiated the second term incorrectly.

You can also solve with some trig/geometry.

Last edited by mqb2766; 4 weeks ago

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(Original post by

Possibly. It's a reasonably common problem, but not worked that particular one through. Upload what you tried if you want it checked. One of the hints is relevant for the chainrule/differentiation part.

**mqb2766**)Possibly. It's a reasonably common problem, but not worked that particular one through. Upload what you tried if you want it checked. One of the hints is relevant for the chainrule/differentiation part.

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#8

(Original post by

Yeah, I used that hint to the best of my ability and thought that it would mean my second term would differentiate to (xv)/(3√(a-x)^2+(a/2)^2), but then when I took the square root part out of the fraction and had it as (3/v)((a-x)^2+(a/2)^2))^1/2, I got a very different term.

**domm1**)Yeah, I used that hint to the best of my ability and thought that it would mean my second term would differentiate to (xv)/(3√(a-x)^2+(a/2)^2), but then when I took the square root part out of the fraction and had it as (3/v)((a-x)^2+(a/2)^2))^1/2, I got a very different term.

time on beach + time in water

...

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(Original post by

Can you upload a pic of what you started with etc? It's got to be

time on beach + time in water

...

**mqb2766**)Can you upload a pic of what you started with etc? It's got to be

time on beach + time in water

...

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#11

(Original post by

But on #1 I mentioned what I started with. For part B, t=(x/v) + (√(a-x)^2+(a/2)^2)/(v/3)

**domm1**)But on #1 I mentioned what I started with. For part B, t=(x/v) + (√(a-x)^2+(a/2)^2)/(v/3)

Sometimes it's easier to see someone's work than to ask about all formatting problems.

Last edited by mqb2766; 4 weeks ago

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**mqb2766**)

I could say the brackets are wrong in the root ...

Sometimes it's easier to see someone's work than to ask about all formatting problems.

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#13

(Original post by

**domm1**)Then you have

x + 3sqrt((x-a)^2 + (a/2))^2)

The chain rule has the numerator for the second terms derivative

3/2 * d(x-a)^2 /dx

can you do that? I've flipped the sign inside the brackets as it's unchanged.

Last edited by mqb2766; 4 weeks ago

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(Original post by

Ok. Multiply through by v. It's a constant so won't affect the minimum time.

Then you have

x + 3sqrt((x-a)^2 + (a/2))^2)

can you do that?

**mqb2766**)Ok. Multiply through by v. It's a constant so won't affect the minimum time.

Then you have

x + 3sqrt((x-a)^2 + (a/2))^2)

**The chain rule has the numerator for the second terms derivative**

3/2 * d(x-a)^2 /dx3/2 * d(x-a)^2 /dx

can you do that?

**I've flipped the sign inside the brackets as it's unchanged.**
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#15

(Original post by

I'm sorry I don't understand the parts I have boldened.

**domm1**)I'm sorry I don't understand the parts I have boldened.

It's slightly easier to differentiate the left expression. Can you do that? It is the ~numerator you get using the chain rule.

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(Original post by

(x-a)^2 = (a-x)^2

It's slightly easier to differentiate the left expression. Can you do that? It is the ~numerator you get using the chain rule.

**mqb2766**)(x-a)^2 = (a-x)^2

It's slightly easier to differentiate the left expression. Can you do that? It is the ~numerator you get using the chain rule.

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#17

(Original post by

d(x-a)^2 / dx = 2x-2a. Sorry that shouldn't of needed repeating, it was just the way it was written that threw me off.

**domm1**)d(x-a)^2 / dx = 2x-2a. Sorry that shouldn't of needed repeating, it was just the way it was written that threw me off.

You divide by 2 because of the sqrt(...) and multiply by 3 because of the velocity.

So the derivative of 3sqrt(..) Is

3(x-a)/sqrt(...)

So you have

0 = 1 + 3(x-a)/sqrt(...)

Are you ok and can you proceed?

Last edited by mqb2766; 4 weeks ago

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(Original post by

Which is 2(x-a).

You divide by 2 because of the sqrt(...) and multiply by 3 because of the velocity.

So the derivative of 3sqrt(..) Is

So you have

0 = x + 3(x-a)/sqrt(...)

Are you ok and can you proceed?

**mqb2766**)Which is 2(x-a).

You divide by 2 because of the sqrt(...) and multiply by 3 because of the velocity.

So the derivative of 3sqrt(..) Is

**3(x-a)/sqrt(...)**So you have

0 = x + 3(x-a)/sqrt(...)

Are you ok and can you proceed?

I've gotten a quadratic, x^2 -40x +410=0 as my equation, however it gives no real roots, I am confused how this is still not correct.

Last edited by domm1; 4 weeks ago

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#19

(Original post by

Edit: Wait, I thought the derivative should be 1 + (Boldened part) as you have the single x term at the start?

I've gotten a quadratic, x^2 -40x +410=0 as my equation, however it gives no real roots, I am confused how this is still not correct.

**domm1**)Edit: Wait, I thought the derivative should be 1 + (Boldened part) as you have the single x term at the start?

I've gotten a quadratic, x^2 -40x +410=0 as my equation, however it gives no real roots, I am confused how this is still not correct.

How did you get that?

Last edited by mqb2766; 4 weeks ago

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then squared everything to get (...) + 9(x-a)^2 (Think this may of been where I went wrong)

Expanded everything out, added like terms and multiplied by 4 just to get rid of the (a^2)/4 fraction to get 40x^2 - 80ax +41a^2=0

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