Question

Implicit differentiation: x^3 + y^3 = 3xy^2

Answer 1

Well just set it up

\[\large \frac{d}{dx} (x^3 + y^3)= \frac{d}{dx} (3xy^2)\]
\[\large \frac{d}{dx} x^3 + \frac{d}{dx} y^3= \frac{d}{dx} 3xy^2\]

What would we do next?

Answer 2

i'm not sure... :(

Answer 3

do we have to get the y's to one side?

Answer 4

Well we have to differentiate first :)

I know you can handle the first term...what is the derivative of x^3 with respect to x?

Answer 5

3x^2 dx

Answer 6

So that's the first term!

\[\large 3x^2 + \frac{d}{dx} y^3 = \frac{d}{dx} 3xy^2\]

Now onto the second term

\[\large \frac{d}{dx} y^3\]

What is this derivative? Hint...use the chain rule

Answer 7

Let me know if you need help there :)

Answer 8

3y^2 dy/dx?

Answer 9

Perfect!!! That's our second term...finish it up for me...what is the final term?

\[\large \frac{d}{dx} 3xy^2\]

Answer 10

3(2y) dy/dx ?

Answer 11

Not quite...we have to use the product rule there
\[\large \frac{d}{dx} 3xy^2 = 3[y^2\frac{d}{dx}x + x\frac{d}{dx}y^2]\]

Answer 12

3y^2 x + xy^2

Answer 13

wait i meant: 3y^2 x + 3xy^2

Answer 14

Not quite *ignoring the 3 to make it look nicer*

\[\large y^2 \frac{d}{dx} x + x \frac{d}{dx} y^2\]
The derivative of x with respect to x is 1
\[\large y^2 + x\frac{d}{dx}y^2\]

Using the chain rule for \(\large \frac{d}{dx}y^2\) what do we get?

Answer 15

2y dy/dx

Answer 16

There we go

So stay with me as I bring all the terms back

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

Everything look good?

Answer 17

i get it!

Answer 18

Perfect...now just solve for \(\large \frac{dy}{dx}\) and you're all set!

Answer 19

could you please wait while i try to get the answer?

Answer 20

Not a problem, post back here when you're done :)

Answer 21

could i simplify 3(y2+2xy dy/dx) to 3y^2 + 6xy dy/dx ?

Answer 22

Absolutely, that is what I would do!

Answer 23

ok! just give me a moment

Answer 24

dy/dx = (3xy^2 - 3x^2)/(3xy) ?

Answer 25

Hmm let's see here

\[\large 3x^2 + 3y^2 \frac{dy}{dx} = 3y^2 + 6xy\frac{dy}{dx}\]

Get the \(\large \frac{dy}{dx}\) to one side and the rest on the other
\[\large 3x^2 - 3y^2 = 6xy\frac{dy}{dx} - 3y^2\frac{dy}{dx}\]

Factor out the common term

\[\large 3x^2 - 3y^2 = \frac{dy}{dx}(6xy - 3y^2)\]
And finally divide both sides by \(\large 6xy - 3y^2\)
\[\large \frac{dy}{dx} = \frac{3x^2 - 3y^2}{6xy - 3y^2}\]

We can even simplify this farther by factoring out a 3 from top and bottom

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

Giving us
\[\large \frac{dy}{dx} = \frac{x^2 - y^2}{2xy - y^2}\]

Answer 26

thank you so much! i see what i did wrong. i had subtracted the 3y^2 dy/dx wrong.

Answer 27

Not a problem! Glad it all made sense!

Answer 28

thanks again!