Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
f(x) = x^3
f'(x) = 3x^2
why then when we use this notation:
y = x^3
dy/dx = 3x^2(dx/dy) <isn't y the dependent variable.. I'm so confused today.
 2 years ago
 2 years ago
f(x) = x^3 f'(x) = 3x^2 why then when we use this notation: y = x^3 dy/dx = 3x^2(dx/dy) <isn't y the dependent variable.. I'm so confused today.
 2 years ago
 2 years ago

This Question is Closed

101Ryan101Best ResponseYou've already chosen the best response.0
I should go study...
 2 years ago

pk51Best ResponseYou've already chosen the best response.1
Two peple developed Calculus, Newton and Leibniz. Newton chose to use primes "'" , or f'(x) while lebinz chose dy/dx, the differential in y/the differential in x. They both mean the same thing for now, but dy/dx becomes important later on in Calculus 2 and 3.
 2 years ago

101Ryan101Best ResponseYou've already chosen the best response.0
when you're doing related rate problems though that extra dx/dt... bllllaaaaa.. :(
 2 years ago

polpakBest ResponseYou've already chosen the best response.2
y is the dependent variable. y' = dy/dx = d/dx of y
 2 years ago

acrossBest ResponseYou've already chosen the best response.0
The notation dy/dx is very useful when integrating and it can simply be read as "the derivative of y with respect to x."
 2 years ago

polpakBest ResponseYou've already chosen the best response.2
you don't get the dx/dy when you use leibniz notation taking the differential of that equation.
 2 years ago

101Ryan101Best ResponseYou've already chosen the best response.0
I get confused when we start doing related rates and dt is thrown in the mix. that what's going on.
 2 years ago

polpakBest ResponseYou've already chosen the best response.2
\[y = x^3\]\[\implies \frac{d}{dx}[y] = \frac{d}{dx}[x^3]\]\[\implies \frac{dy}{dx} = 3x^2\]
 2 years ago

101Ryan101Best ResponseYou've already chosen the best response.0
Ok, and so if I took both sides with respect to dt... that's why there's an "extra" dx/dt.. Ok... I get it.
 2 years ago

polpakBest ResponseYou've already chosen the best response.2
If however we wanted to take the differential with respect to t: \[\frac{d}{dt}[y] = \frac{d}{dt}[x^3]\]\[\implies \frac{dy}{dt} = \frac{d}{dx}[x^3] \cdot \frac{dx}{dt}\]\[\implies \frac{dy}{dt} = 3x^2\frac{dx}{dt}\]
 2 years ago

polpakBest ResponseYou've already chosen the best response.2
I <3 Leibniz notation. Newton can suck it.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.