Confusing Differentiation

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

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

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.

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

https://isaacphysics.org/questions/lifeguard_num?board=4cb67965-7420-465a-8aba-1ea18244028f

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

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.

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.

Can you upload a pic of what you started with etc? It's got to be
time on beach + time in water
...

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

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.

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

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

(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.

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.

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?

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.

Yes 1+3(x-a)/sqrt(..) (typo).
How did you get that?