Find dy/dx in terms of x and y if x^(1/3) = 5sqrt(y)
I don't understand this question. Is this supposed to be a simple derivative or is it different?

Question

Find dy/dx in terms of x and y if  x^(1/3) = 5sqrt(y)
I don't understand this question. Is this supposed to be a simple derivative or is it different?

australopithecus · Answer

This is just explicit differentiation

australopithecus · Answer

You can just solve for y if you want then take the derivative, as dy/dx y is y'

Or for example you can do this,

dy/dx x^2 + y^2 = dy/dx 1

is

2x + 2yy' = 0
solve for y'
y' = -2x/2y

australopithecus · Answer

when differentiating y you need to use chain rule on it, that is where y' comes from because y is the function itself

australopithecus · Answer

or rather y is a function

australopithecus · Answer

so for your problem,

x^(1/3) = 5(y)^(1/2)

take the derivative of both sides

and solve for y'

australopithecus · Answer

Ignore this "You can just solve for y if you want then take the derivative, as dy/dx y is y'."

best to just use the method I showed you

australopithecus · Answer

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

anonymous · Answer

so the answer is (1/3)x^(2/3)-( 5(1/2)y(-1/2) ) = dy/dx ?

australopithecus · Answer

you took the derivative incorrectly,
 
Also you messed up the algebra, use y' do not use dy/dx