Question

what is the derivative of x/y with respect to x?

Answer 1

1/y, if i'm not wrong

Answer 2

if it was with respect to y, would it be 1/yprime?

Answer 3

no, then it would be x

Answer 4

are you sure? i'm going by this example from class, where prof said derivative of \[x^2 + y^2\] was 2x + y'

Answer 5

sorry, i meant 2x + 2yy'*

Answer 6

He is differentitating the expression implicitly with respect to x.

Answer 7

going by that example, i actually need a y' in my original question(i think its respect to y instead of x).

Answer 8

y is a function of x.
The derivative of y with respect to x is dy/dx, also written as y'.

Answer 9

yeah, there implicit and explicit differentiation and if your prof is differentiating implicitly it's right. i was thinking of explicit diff.

Answer 10

When you take a derivative this way, you still have to use the power rule, product rule, etc. and the chain rule.

Answer 11

yea, in the class example, it was easier to differentiate since they were on different terms. in this one, x and y are in the same one term. i'm not sure how do i approach? i haven't done this in a while.

Answer 12

@mathstudent55 so i use a quotient rule here where i treat both variables as non-constants?

Answer 13

You have two terms.
x^2 and y^2

Answer 14

The derivative of x with respect to x is 1.
The derivative of x^2 using the power rule is simply 2x.
You do apply the chain rule, but since the derivative of x with respect to x is 1,
the derivative of x^2 = 2x * 1 = 2x

Answer 15

Now let's deal with the other term, y^2.
The derivative of y with respect to x is y'.
When you differentiate y^2, youi apply the power rukle firstr followed by the chain rukle.

\(\dfrac{d}{dx} y^2 = 2y \times \dfrac{d}{dx}y = 2yy'\)

Answer 16

The power rule:

Function u is a function of x, then

\(\dfrac{d}{dx} u^n = n u^{n -1 }\dfrac{d}{dx} u\)

Answer 17

you could use the quotient rule or write it as
\[ x \ y^{-1} \]
and use the product rule
\[ \frac{d}{dx} x \ y^{-1} = x \frac{d}{dx} y^{-1} +y^{-1} \frac{d}{dx} x \]
you get
\[ x \cdot -y^{-2} \frac{dy}{dx} + y^{-1} \frac{dx}{dx} \\
= -x\frac{y'}{y^2} + \frac{1}{y}
\]

Answer 18

ty