Question

Find the derivative of the function y defined implicitly in terms of x
y^2 = ln(3x+1y)

Answer 1

i'm confused about the explicitly in terms of x

Answer 2

explicitly is in y= some equation with "x"

Answer 3

implicitly you take the derivative of the entire equation

Answer 4

\[\frac{d}{dx}[y^2] = 2y(\frac{dy}{dx})\]
are you familiar with this?

Answer 5

yes
but i must have done it wrong
I got 3/ (6xy-y+2y^2)

Answer 6

\[\frac{d}{dx}[\ln(u)] = \frac{u'}{u}\] did you use this rule as well?

Answer 7

for ln, i used 1/x times derivative of x

Answer 8

oh i got it :)
but thanks for clarifying the implicit in terms of y thing. i interpreted it as d/dx

Answer 9

so how woudl you know it was d/dx or d/dy and what does that mean?

Answer 10

yes, same thing,
\[\frac{ 1 }{ x } * x' = \frac {x'}{x}\]

let's see if i can do this. we got the left part:
\[2y\frac{dy}{dx} = \frac{3 + \frac{dy}{dx}}{3x+y}\]

Answer 11

d/dx is the derivative with respect to x and d/dy is the derivative with respect to y

Answer 12

here's a few examples:

Answer 13

\[\frac{d}{dx}[x] = 1\]
\[\frac{d}{dx}[y] = \frac{dy}{dx}\]
\[\frac{d}{dt}[x] = \frac{dx}{dt}\]
\[\frac{d}{dt}[y] = \frac{dy}{dt}\]

Answer 14

basically the d/d means "the derivative of" and "y" in the numerator means "the derivative of y" and an "x" in the denominator (dy/dx) means "the derivative of y with respect to x"

Answer 15

what about dy/dx [y] or does that happen?

Answer 16

dy/dx[y] = 0 because there is no X variables to respect. we assume all other variables are constant

Answer 17

oh ok. Thank you!!

Answer 18

i dont think "y" can be in the derivative operator but if it was y = h^2 and dy/dx[y] then it's 0

Answer 19

since y is a function

Answer 20

or equation or w/e

Answer 21

y^2 = ln(3x+1y)
2y dy = 3 dx/(3x+1)
dy/dx=3/(2y*(3x+1))
y'=3/(2y*(3x+1))
y'=3/(2sqrt(ln(3x+1))*(3x+1))

Answer 22

thanks to you too!

Answer 23

(3x+y) i mean * lol