Question

Differentiate Implicitly.
x+xy-2x^3=2

Can someone please show me how?

Answer 1

@eggshell
do you know how to differentiate x and y with respect to x?

Answer 2

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

Answer 3

basically if you know this you are golden
plus all the rules like product, sum, constant multiply rule, constant rule, and chain rule, quotient rule...

Answer 4

\[\frac{d}{dx}(x+xy-2x^3)=\frac{d}{dx}(2)\]
on the left use sum rule
on the right use constant rule
that is
\[\frac{d}{dx}(x)+\frac{d}{dx}(xy)-\frac{d}{dx}(2x^3)=0\]

Answer 5

so can you tell me which of the 3 derivatives there you are having issues with?

Answer 6

@freckles yes I know how to differentiate but not the xy

Answer 7

x*y is a product

Answer 8

you need to use product rule

Answer 9

\[\frac{d}{dx}(xy)=y \frac{d}{dx}x+x \frac{d}{dx}y\]

Answer 10

Thank you!

Answer 11

wait but that gives me x+y so it gives me (1+x+y-6x^2) the book says the answer is (6x^2-y-1)/x

Answer 12

no no...

Answer 13

did you use this:
\[\frac{d}{dx}x=\frac{dx}{dx}=1 \\ \frac{d}{dx}y=\frac{dy}{dx} =y' \]

Answer 14

\[\frac{d}{dx}(xy)=y \frac{dx}{dx}+x \frac{dy}{dx}=y \cdot 1 +x \cdot y'=y+xy' \]

Answer 15

\[\frac{d}{dx}(x)+\frac{d}{dx}(xy)-\frac{d}{dx}(2x^3)=0 \\ 1+y+xy'-6x^2=0\]
now you solve that for y'

Answer 16

the d/dx and dy/dx thing confuse can you explain the difference?

Answer 17

d/dx means we are taking the derivative with respect to x
dy/dx means we have taken the derivative of y with respect to x

Answer 18

like you don't write d/dx by itself
like you have to have d/dx (some function) in order to take derivative of that function with respect to x

Answer 19

d/dx (y) =dy/dx=y'

Answer 20

maybe you prefer that latter notation

(xy)'=(x)'y+x(y)'=1y+xy'
since (x)'=1 and (y)'=y'

Answer 21

No I understand it but isn't y' just one because you bring down the exponent? Or should I just leave it as y' when doing implicit differentiation?

Answer 22

and what does with respect to x mean?

Answer 23

you put y' when you find the derivative of y with respect to x

Answer 24

and no y' doesn't equal 1 unles y=x

Answer 25

but we are not given y=x

Answer 26

y is a function of x

Answer 27

But why does this example make it look like its the exact same thing as in derive both x and y exactly the same

Answer 28

they used chain rule for y part and power rule for x part

Answer 29

they used that derivative wrt to x of y is y'

Answer 30

they are treating y and x no different than i have

Answer 31

\[\frac{d}{dx}(4x^2)=4 \cdot 2 x^{2-1} \frac{dx}{dx}=8x(1)=8x \\ \frac{d}{dx}(-2y^2)=-2 \cdot 2y^{2-1} \frac{dy}{dx}=-4y \frac{dy}{dx}\]

Answer 32

and I have said here:
\[\frac{d}{dx}(xy)=y \cdot \frac{dx}{dx}+x \cdot \frac{dy}{dx} \text{ by product rule } \\ \frac{d}{dx}(xy)=y \cdot 1 +x \frac{dy}{dx}=y+x \frac{dy}{dx}\]

Answer 33

just like when i differentiated y above i wrote y'

Answer 34

shouldn't be any different when differentiate y elsewhere

Answer 35

ahhh I get it now ! it says to also solve the equation for y as a function of x and to find dy/dx. what is that asking em to do?

Answer 36

\[1+y+xy'-6x^2=0\]
it wants you to solve this for y'
put everything on the other side by addition or subtraction (except the xy' term)
then divide both sides by x.

Answer 37

this will give you dy/dx as a function of x

Answer 38

oh and it looks like they also want you to solve your first equation for y too

Answer 39

how would I do that second part?