Question

DE

Answer 1

Find dy/dx
\[x^{3} y^{4} = (x-y)^{7} \]

Answer 2

How I did:
\[\large x^{3} 4 y^{3} + y^{4} 3 x^{2} = 7(x-y)^{6} \times (1-\frac{dy}{dx}) \]

Answer 3

is it correct?what after this?

Answer 4

@hba

Answer 5

\[x ^{2}y ^{3}\]should be attached a dy/dx since you found the derivative of y there.

Answer 6

where?

Answer 7

First step on the left hand side of equation, during the u'v + uv' process, at the point you do \[4y ^{3}\]you shoud add dy/dx\[x ^{3}4y ^{3}\frac{ dy }{ dx }\]

Answer 8

wont it be like this?
\[\large x^{3} y^{4} = y{4} \frac{dx^{3}}{dx} \times x^{3}\frac{dy^{4}}{dx}\]

Answer 9

^^ yes but when you differentiate "y^4" you apply the chain rule since it is wrt x the derivative of y is dy/dx

...hence (4y^3)* dy/dx

Answer 10

why wont we do that for x?

Answer 11

technically yes, however dx/dx is just 1

Answer 12

okay so the equation becomes

Answer 13

maybe this will make more sense...
\[\frac{dy^{4}}{dx} = \frac{dy^{4}}{dy}*\frac{dy}{dx}\]

Answer 14

why did u write this :o

Answer 15

to show what is going on...you are taking derivative of y^4 and getting 4y^3 right? but that is wrt y since you want it wrt x , you must multiply by dy/dx

Answer 16

'but that is wrt y"
?

Answer 17

\[\frac{d y^{4}}{dx} \]

Answer 18

4y^3

Answer 19

with respect to y, ...y is the variable

for the (1-dy/dx) part which is correct.....you said this is derivative of (x-y) right
well derivative of y with respect to y is 1 but then you multiply by dy/dx

Answer 20

in general, whenever you differentiate a term with variable other than "x" you must multiply by dy/dx

Answer 21

makes sense :D

Answer 22

\[\frac{dz^{2}}{dx} = 2z*\frac{dz}{dx}\]

Answer 23

how will we simplify this big equation last question :o

Answer 24

little algebra :)
its all about isolating the "dy/dx" term

Answer 25

okay!

Answer 26

distribute the RHS .... then move "dy/dx" terms to 1 side .... factor out dy/dx.... divide

Answer 27

"in general, whenever you differentiate a term with variable other than "x" you must multiply by dy/dx"

little confusing still..that example u gave isnt 100% clear,but i just have an idea

Answer 28

ok example was z^2 ..... if it was x^2 then derivative is 2x right?
in other words
\[\frac{dx^{2}}{dx} = 2x\]
so it follows
\[\frac{dz^{2}}{dz} = 2z\]
thus
\[\frac{dz^{2}}{dx} = \frac{dz^{2}}{dz}*\frac{dz}{dx} = 2z*\frac{dz}{dx}\]
**notice the "dz" cancels so the equality holds

Answer 29

thanks!

Answer 30

yw

here is a good reference for future
http://tutorial.math.lamar.edu/Classes/CalcI/ImplicitDIff.aspx

Answer 31

would be helpful!!

Answer 32

i got the answer as
-3y/4x

Answer 33

\[\large y^{4} \frac{(dx^{3})}{dx} +x^{3} \frac{d y^{4}}{dy} \times \frac{dy}{dx} = 7(x-y)^{6} \times (1-\frac{dy}{dx}) \]

Answer 34

\[\large y^{4}3x^{2} +^{3}4y^{3} \frac{dy}{dx} = 7(x-y)^{6} (1-\frac{dy}{dx})\]

Answer 35

\[ y^{4}3x^{2} = 7(x-y)^{6} - 7(x-y)^6 \frac{dy}{dx}-x^{3}4y^{3}\frac{dy}{dx}\]

Answer 36

\[\large \frac{dy}{dx}(7(x-y)^{6} - 7(x-y)^{6} -x^{3}4y^{3})=y^{4} 3x^{2} \]

Answer 37

\[\large \frac{dy}{dx}(-x^{3}4y^{3}) = y^{4}3x^{2}\]

Answer 38

\[\large \frac{dy}{dx}= \frac{y^{4}3x^{3}}{-x^{3}4y^{3}} \]

Answer 39

\[\large \frac {dy}{dx}=-\frac{3y}{4x}\]

Answer 40

how far is this correct?

Answer 41

2nd last step==>
thats x^2 not x^3

Answer 42

this is incorrect
\(\large \frac{dy}{dx}(7(x-y)^{6} - 7(x-y)^{6} -x^{3}4y^{3})=y^{4} 3x^{2}\)

Answer 43

why?

Answer 44

oh okay

Answer 45

then what wud be the next step?

Answer 46

factor out dy/dx
there r only 2 terms with dy/dx not 3

Answer 47

continue from here
\(y^{4}3x^{2} = 7(x-y)^{6} - 7(x-y)^6 \frac{dy}{dx}-x^{3}4y^{3}\frac{dy}{dx}\)

Answer 48

only last 2 terms has dy/dx

Answer 49

\[\frac{dy}{dx}= (7(x-y)^{6} +x^{3}y^{4}) = 7(x-y)^{6}-y^{4} 3x^{2})\]

Answer 50

sorry the equal to isnt there wth!

Answer 51

dy/dx times* not equals

Answer 52

why x^3 y^4 ??

Answer 53

x^3 4y^3

Answer 54

typo error!!

Answer 55

then its correct...u can solve further...just divide.

Answer 56

yayyy!

Answer 57

thanks!