Main content

## Differential Calculus

### Course: Differential Calculus > Unit 4

Lesson 4: Introduction to related rates- Related rates intro
- Analyzing problems involving related rates
- Analyzing related rates problems: expressions
- Analyzing related rates problems: expressions
- Analyzing related rates problems: equations (Pythagoras)
- Analyzing related rates problems: equations (trig)
- Analyzing related rates problems: equations
- Differentiating related functions intro
- Worked example: Differentiating related functions
- Differentiate related functions

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Differentiating related functions intro

Sometimes we have an equation that relates to functions of the same variable. Using the chain rule, we can find the derivatives of those functions with respect to that variable.

## Want to join the conversation?

- I'm so confused, where did t come from? How do we know x and y are functions of t?(12 votes)
- 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.(3 votes)

- Could you also solve this equation using implicit differentiation?(8 votes)
- 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.(2 votes)

- I don't understand how dy/dx becomes (1/2)x^(-1/2) instead of 1/sqrtx(3 votes)
- 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.(4 votes)

- 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.(2 votes)- 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.(5 votes)

- 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(2 votes)- 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(4 votes)

- 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?(2 votes)- 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.(2 votes)

- 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.(1 vote)
- 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.(2 votes)

- I got different answer somehow.

dx/dt = 12, get x(t)=12t,

then when x = 9, t is 3/4.

y = root x, so y is root of 12t,

so dy/dt is 1/2*(12t)^(-1/2), plug in t=3/4, the answer is 1/6.

Please tell me where I did wrong.(1 vote)- 𝑑𝑥∕𝑑𝑡 = 12 ⇒ 𝑥 = 12𝑡 + 𝐶

(𝐶 will cancel out with itself later on, but for now we should have it there)

𝑥 = 9 ⇒ 12𝑡 + 𝐶 = 9 ⇒ 𝑡 = (9 − 𝐶)∕12

𝑦 = √𝑥 = √(12𝑡 + 𝐶)

⇒ 𝑑𝑦∕𝑑𝑡 = 1∕(2√(12𝑡 + 𝐶))⋅12 (chain rule!)

= 6∕√(12𝑡 + 𝐶)

Plugging 𝑡 = (9 − 𝐶)∕12, we get

𝑑𝑦∕𝑑𝑡 = 6∕√(12(9 − 𝐶)∕12 + 𝐶) = 6∕√(9 − 𝐶 + 𝐶) = 6∕√9 = 6∕3 = 2(1 vote)

- I just got confused ! dy/dt = dy/dx.dx/dt right ! and dx/dt = 12 ! plus y= square root of x which means dy/dx = dx/dt over the root of x * 2, which leads to dy/dt = 1/6 am I wrong ?(1 vote)
- I"m not sure how you got a link into your question, but if I'm understanding your question then you have made a mistake.

The mistake is that you have ignored the**value**of dx/dt ...(1 vote)

- What does "x=f(t)" look like when it's graphed? I tried on Desmos but I just get an error. Is t one of the axes on the plane? If so, which one? Thanks.(1 vote)
- We don't know its graph because x(t) isn't explicitly defined, but you could draw a graph for it on the x-t plane if the equation is given.

In most cases, you won't be asked to graph it. t here can be considered as a parameter, which is something both x and y depend on. Most people consider it as time, and x(t) and y(t) are seen as the x and y coordinate of a particle w.r.t time. So, on an x-y plane, you can't actually represent t. Instead, you can take x(t) and y(t) and plot points for different t values. The plot you get is the path the particle traces as time increases.(1 vote)

## Video transcript

- [Instructor] We are told
the differentiable functions X and Y are related by
the following equation, Y is equal to square root of X. And it's interesting, they're telling us that they are both
differentiable functions, even X is a function, must be
a function of something else. Well, they tell us that
the derivative of X with respect to T is 12 and they want us to
find the derivative of Y with respect to T when X is equal to nine. So, let's just make sure
we can understand this. So, they're telling us that
both X and Y are functions. Arguably they are both functions of T. Y is a function of X but then X is a function of T so Y could also be a function of T. One way to think about it is if X is equal to F of T, then Y is equal to the square root of X which would just be F of T. Another way to think about it if you took T as your input into your function F, you're going to produce X and then if you took that as your input into the square root function you are going to produce Y. So, you could just view this
as just one big box here, that Y is a function of Y but now let's actually
answer their question. To tackle it we just have
to apply the chain rule. The chain rule tells us
that the derivative of Y with respect to T is going to be equal
to the derivative of Y with respect to X times
the derivative of X with respect to T. So, let's apply it to
this particular situation. We're gonna have the derivative of Y with respect to T is equal to the derivative of Y with respect to X. Well, what's that? Well, Y is equal to the
principle root of X. You could also write this as Y is equal to X to the one half power. We could just use the power rule. The derivative of Y with respect to X is one half X to the negative one half. So, let me write that down. One half X to the negative one half and then times the derivative of X with respect to T times
the derivative of X with respect to T. Well, let's see, we wanna find
what we have here in orange, that's what the questions asks us. They tell us when X is equal to nine and the derivative of X with respect to T is equal to 12. So, we have all of the information necessary to solve for this. So, this is going to be equal to one half times nine to the negative one half, nine to the negative one half, times DX/DT, the derivative
of X with respect to T is equal to 12, times 12. So, let's see, nine to the one half would be three, nine to the negative
one half would be 1/3, so this is 1/3, so this will all simplify
to 1/2 times 1/3 is 1/6, so we could have a six in the denominator and then we are going to
have a 12 in the numerator. So, 12/6, so the derivative
of Y with respect to T when X is equal to nine and derivative of X with
respect to 12 is two.