Question

find the derivative of...

(x^2)(y)+(x)(y^2)=6

Answer 1

\[x ^{2}y+xy ^{2}=6\]

Answer 2

with respect to? x or y?

Answer 3

find: dy/dx

Answer 4

2xy+y^2 = 0

Answer 5

how did you do it?

Answer 6

\[x ^{2}y+xy ^{2}=6\]
\[2xy+x^2y'+y^2+2xyy'=0\] solve for \(y'\)

Answer 7

you are thinking of \(y=f(x),y'=f'(x)\) so you need the product rule for both parts

Answer 8

\[x ^{2}+2xy+2xy+y ^{2}=0\]

Answer 9

thats what i got using the product rule

Answer 10

you also need the chain rule because \(y\) is a function of \(x\)
so the derivative of \(y^2\) is \(2yy'\) as the derivative of \(f^2(x)\) is \(2f(x)f'(x)\)

Answer 11

i still need help guys i dont get it

Answer 12

do know about product rule?

Answer 13

yes i know all the rules i just can't figure out this specific problem

Answer 14

\[\frac{ d }{ dx }(x ^{2}y + x y^{2}) = 0\]
now apply product rule

Answer 15

\[x ^{2}+2xy+2xy+y ^{2}=)\]

is that correct?

Answer 16

=0

Answer 17

it should be like
\[x ^{2}y' + 2xy + y ^{2}+ 2xy' = 0\]

Answer 18

notice
\[ \frac{d}{dx} y= \frac{dy}{dx} \]

Answer 19

we can also write
\[\frac{ dy }{ dx } = y'\]

Answer 20

as compared to
\[ \frac{d}{dx} x= \frac{dx}{dx}=1\]

Answer 21

im so confused
i honestly don't get this dx/dy or d/dy stuff

Answer 22

first you have to know about you are doing derivative to w.r.t to x or y

Answer 23

If you need more info on derivatives, these might help
http://www.khanacademy.org/math/calculus/differential-calculus/v/calculus--derivatives-1
when you get time.

meanwhile, I assume you know how to take the derivative wrt x of
\[ \frac{d}{dx} (y=x)\]
you get
\[ \frac{d}{dx} y=\frac{d}{dx} x\]
\[ \frac{dy}{dx} =\frac{dx}{dx} \]
\[ \frac{dy}{dx} =1 \]
or this problem
\[ \frac{d}{dx} (y=x^2)\]
you get
\[ \frac{dy}{dx} =\frac{d}{dx}x^2 \]
\[ \frac{dy}{dx} =2x\frac{dx}{dx} \]
\[ \frac{dy}{dx} =2x \]

Answer 24

now for a product rule problem
d (uv) = u dv + v du


\[ \frac{d}{dx} x^2 y = x^2 \frac{d}{dx}y + y \frac{d}{dx}x^2 \]
\[ \frac{d}{dx} x^2 y = x^2 \frac{dy}{dx} + y 2x\frac{dx}{dx} \]
\[ = x^2 \frac{dy}{dx} + 2xy \]