Question

implicity derive - 7x^3+6y^4 = 21

Answer 1

\[7x ^{3}+6y ^{4} =21\]

Answer 2

how would you type this in wolfram

Answer 3

derive it as usual then solve for y'

Answer 4

d/dx (f(x,y))

Answer 5

= sign confusing me

Answer 6

you dont derive an = sign :)

Answer 7

haha i know just confusing me

Answer 8

7x^3+6y^4 = 21

d(7x^3)/dx = ?

Answer 9

21x^2 +24y^3 * dy/dx = 0
now make dy/dx the subject

Answer 10

6y^4

Answer 11

what about the 21

Answer 12

we can get to that ...
\[\frac{d(7x^3)}{dx} = ?\]

Answer 13

7x^2

Answer 14

the derivative of 21 is 0

Answer 15

would i just write d/dx 21

Answer 16

why do people say derive when they mean differentiate?

Answer 17

not quite, and you are missing a part:

\[\frac{d(7x^3)}{dx} = 21x^2\frac{dx}{dx}\]

Answer 18

same thing

Answer 19

what is this now:

\[\frac{d(6y^4)}{dx}=?\]

Answer 20

they are different

Answer 21

confused

Answer 22

it drives me crazy a little lol

Answer 23

minds are well tuned then =] how to solve this???

Answer 24

\[\frac{d(6y^4)}{dx}=?\]

Answer 25

mistre confused here

Answer 26

youll have to be more specific, I can quite read your mind to see what you are actually confused about ...

Answer 27

not sure wether its just d6y4/dx

Answer 28

would it be more solvable for you if it were like this?

\[\frac{d(6x^4)}{dx}=?\]

Answer 29

not sure how to multiply that out

Answer 30

how would you write \[\frac{d}{dx}6[f(x)]^4\]

Answer 31

if you have h(x)=6(f(x))^4 and i asked you to find h' what would you say?
hopefully you know to use chain rule and say
h'(x)=6*4*(f(x))^3*f'(x)=24f(x)[f(x)]^3

Answer 32

lol zarkon thats what i was gonna say or did

Answer 33

yes now its just d5x^4

Answer 34

nice ;)

Answer 35

oops i missed my little prime thingy

Answer 36

only on the second part ;)

Answer 37

the rules for differentation dont change becuase the variable changes;

6x^4 derives in the exact same manner as 6y^4 right?

Answer 38

h'(x)=24f'(x)[f(x)]^3

Answer 39

thats true but dx/dy in implicit differentiation confuses me

Answer 40

7x^3+6y^4=21 how to solve this one i have 2 more to post

Answer 41

consider this then:
\[\frac{d}{dx}(6x^4)=\frac{dx}{dx}(24x^3)\]
out poped the variable, do you see it? happens all the time; but the thing is dx/dx = 1 so they toss it out.
\[\frac{d}{dx}(6y^4)=\frac{dy}{dx}(24y^3)\]
the variable still pops out of it, but we have to keep this derived bit left in the equation since it has no "real" value to associate it to

Answer 42

ok understood that

Answer 43

good :)

and the 21 on the other side is just a constant; and all constant derive to 0

Answer 44

so now how to write the whole thing out ?

Answer 45

7x^3 +6y^4 = 21 ; derives to

\[21x^2\frac{dx}{dx} + 24y^3 \frac{dy}{dx} = 0\]
\[21x^2 + 24y^3 \frac{dy}{dx} = 0\]
now we solve for dy/dx

Answer 46

wouldnt it be 21x^3 or am i forgetting dx/dx

Answer 47

- 21x^2 and divide out the 24y^3 right?
\[\frac{dy}{dx}=\frac{-21x^2}{24y^3}\]

Answer 48

right

Answer 49

7x^3 just derives to:

(3) 7x^2 (dx/dx) = 21x^2

Answer 50

wouldnt 24 be on top half of the equation for that example

Answer 51

no, the rest of it is just algebra ....

ax + by = 0

by = -ax

y = -ax/b

Answer 52

right sorry about that

Answer 53

will start a new thread