Question

Calculate y'
(x)(y^4)+(x^2)(y)=(x)+(3)(y)

Answer 1

take derivative of both sides. once you are done, isolate y'.

Answer 2

how exactly am i supposed to derive it

Answer 3

i know it has something to do with the implicit derivative

Answer 4

let start with each terms.

\(\dfrac{d}{dx} xy^4 = ?\)

Answer 5

yeah it is implicit derivative

Answer 6

y^4+x*4y^3

Answer 7

hmm you forgot something.

Answer 8

remember? \(\dfrac{d}{dx}y = y'\) right?

so it's supposed to be \(y^4 y' + 4xy^3 y'\)

Answer 9

does that make sense?

Answer 10

hmm, why is it y' and not x' ?

Answer 11

because you take derivative in respect to x. so x' = d/dx x = 1

Answer 12

y' because we don't know what y is so we just let it y'. y is function of x.

Answer 13

is that clear?

Answer 14

yes

Answer 15

ok now next term in left side.
\(\dfrac{d}{dx} x^2y = ?\)

Answer 16

y'*2x+y'*x^2

Answer 17

close. it's \( 2xy + x^2y'\)

Answer 18

when you take derivative of x, you dont do it to y. get it?

Answer 19

but in x*y^4 you took the d/dx of x and got 1

Answer 20

\(\dfrac{d}{dx}uv = u'v + uv'\)
u is x^2 and v is y. we take derivative of x^2, so we need to left y untouched.

yeah d/dx of x is 1 so d/dx x*y^4 = 1*y^4 + 4 x y^3 y'

Answer 21

so is that clear??

Answer 22

so i'm not quite sure when you have y'...

Answer 23

you get y' when you take derivative of y.

Answer 24

so for example:
d/dx y^2 = 2y^1 y'
because of chain rule; we just deal with exponent, then we deal with y itself.

Answer 25

when we took the d/dx of x*y^4, i was supposed to use the chain rule?

Answer 26

and product rule. we use product rule first

Answer 27

for y part, we then use chain rule

Answer 28

ok but we have to take the d/dx of x and you said to put y'? but we didnt take the d/dx of y tho

Answer 29

yeah, because we left y untouched.

\(\dfrac{d}{dx} xy^4 = \left( \dfrac{d}{dx}x\right)y^4 + x\left(\dfrac{d}{dx}y^4\right)\)

\( = 1\cdot y^4 + x\cdot 4y^3 \space y'\)

Answer 30

when we take derivative of y^4, we used chain rule

Answer 31

ohh...

Answer 32

so you understand?

Answer 33

yes kind of

Answer 34

ok try again.
d/dx x^2 y

Answer 35

so d/dx of x is 1 and d/dx of y =y' ?

Answer 36

yeah

Answer 37

we take derivative of y with respect to x. we don't know what y is equal to so we just left it y', which is another way to say d/dx y

Answer 38

d/dx of x^2 y = (2xy) + (x^2)(y')

Answer 39

yeah!

Answer 40

ok

Answer 41

now take derivative of right side

Answer 42

d/dx (x) + 3y' ?

Answer 43

i mean 1 + 3 y'

Answer 44

yeah, 1 + 3y'

Answer 45

so all together, we get
\(y^4 + 4xy^3y' + 2xy + x^2y' = 1+3y'\)

Answer 46

now we are done with derivative. left thing for us to do is to isolate y'

Answer 47

can you do that?

Answer 48

y'(4xy^3+x^2-3)+y^4+2xy-1 ?

Answer 49

y'=(4xy^3+x^2-3)/-(y^4+2xy-) ?

Answer 50

-(y^4+2xy+1)

Answer 51

yeah, now distribute negative sign in denominator, then that's your final answer.

Answer 52

-(y^4+2xy-1)

Answer 53

alright i get it now, tyvm :P

Answer 54

no problem. glad i helped