Question 1

I'm so confused, where did t come from? How do we know x and y are functions of t?

Accepted Answer

We know because there are derivatives of x and y with respect to t, if they weren't functions of t, then such derivatives would have no value.

Question 2

Could you also solve this equation using implicit differentiation?

Accepted Answer

Using implicit differentiation:
y=sqrt(x)

Take the derivative of both sides (note that we are taking dy/dt, not dy/dx, because we are taking the derivative in terms of t as the question calls for):
dy/dt = (1/2 x^(-1/2))(12)
where (1/2 x^(-1/2)) is dy/dx and 12 is, as given, dx/dt.

When dy/dx is multiplied with dx/dt, we get dy/dt.

Since we are finding dy/dx when x is 9, we get:
dy/dt = (1/2 (sqrt9)^(-1/2))(12)
dy/dt = (1/2 * 1/3)(12)
dy/dt = (1/6)(12)
dy/dt = 2

Basically, this is what is shown in the video, but in a more direct format.

Question 3

I don't understand how dy/dx becomes (1/2)x^(-1/2) instead of 1/sqrtx

Accepted Answer

sqrt(x) can be written as x^(1/2)
d/dx x^n = n*x^(n-1).  Here n =1/2 so let's plug in.

(1/2)*x^(-1/2)  negative exponents can be written as x^(-a) = 1/(x^a) so we can rewrite the answer we got as this

(1/2)*1/(x^(1/2))  then as we said before x^1/2 can be written as sqrt(x)

(1/2)*1/sqrt(x) then we just multiply the fractions

1/(2sqrt(x))

so 1/sqrt(x) is actually pretty close, just need to keep in mind it is also being multiplied by 1/2

Let me know if that did not help.

Question 4

Where did the dx/dt come from? How does that relate to the chain rule?

If we did the chain rule on sqrt(x), that would come out as 1x^-1/2 and that would be it, wouldn't it? Why tack on this extra dx/dt and why is it there?

Does it have something to do with time being a factor in this equation?

Accepted Answer

Sal mentions that the problem states that x AND y are differentiable funtions, so x is also a differentiable function, which means x is a function.  the problem then says dx/dt is 12 so that is basically giving us the answer that x's independent variable is t.  so you can think of y as y(x) or y of x and x as x(t) or x of t.

it relates to the chain rule because we are saying y is a function of  and x is a function of t so it is one function inside of another, which is what the chain rule deals with.

We don't use the chain rule here as we normally do.  instead we fill in the blanks.  the chain rule says dy/dt = dy/dx * dx/dt.  we don't know what x(t) is though so we cannot solve for dx/dt, but the problem gives us what it is so we can just fill it in.

time being a factor is often why you would have a problem like this, but not always, the big takeaway is just how to take derivatives of composite functions that are not written out in the standard way.

Let me know if something didn't make sense, or yous till have questons.

Question 5

Can someone tell me if I understand this correctly in regards to dx/dt.  After re-watching the video I am less clear than a I was a few minutes ago.

dx/dt is saying what is the derivative for the FUNCTION x when you insert the PARAMETER t.

Thanks for the help.

Accepted Answer

dx/dt is an expression meaning "the derivative of x with respect to t". So it's as you said: it's the derivative of the function x when using the variable t.

Question 6

Shouldn't we have 2answers for this question i.e. 2, -2
Explanation:

we want to calculate dy/dt for x= 9 and we know x-y relation so we get y = +3,-3 for which we have to calculate dy/dt

since y = x^.5   ,    so x= y^2

given is,    dx/dt  = 12     
we substitute x with y^2 so above equation becomes  d(y^2)/dt = 12   
so, applying chain rule and simplifying we get,
dy/dt =  6/y

substitute two values of y( which we found at top) in this equation. we get 
dy/dt = +2,-2

Accepted Answer

If you are given a square root, you should presume it to be the principal (i.e. positive) square root.  If you introduce a square root, you need to consider both the positive and negative roots. 
 You might want to review the material on radicals:
https://www.khanacademy.org/math/algebra/rational-exponents-and-radicals/alg1-radicals/v/introduction-to-square-roots

Question 7

i used implicit differentiation and solved the problem. i don't get what's new and important being taught that we need to be concerned about learning in this video.

Accepted Answer

Nothing new then. Assuming you've differentiated w.r.t t, that's what they want you to do here. They want you to get accustomed to having independent variables that aren't x and y.

Question 8

Is this just the chain rule? In the video dy/dt is defined as dy/dx * dx/dt. The chain rule for dy/dx is defined as dy/du * du/dx, where u is the inner function of y. Not sure if there is a gap in my understanding.

Accepted Answer

Yes, this is an application of the chain rule via implicit differentiation.  You may want to review implicit differentiation.  Khan Academy has a lesson on it. The other area where you may have a misunderstanding is in the application of the symbols. The chain rule does not require the outside function be set to y, the inside function be set to u, and that the independent variable (i.e. the argument) of the inside function, u, be set to x. The chain rule only provides a method for finding the derivative of a composite function. The functions of the composite can use any symbols that they wish or that you wish to assign to them.

Course: AP®︎/College Calculus AB > Unit 4

Differentiating related functions intro

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