1. ## Another Implicit differenriarion

Lets see if I learned from the last one.

$\displaystyle 2x^2y=sin2x$

$\displaystyle \frac{d}{dx}[2x^2y] = \frac{d}{dx}[sin2x]$

$\displaystyle 4xy\frac{dy}{dx}[2x^2]=\frac{dy}{dx}[cos2x]\frac{dy}{dx}[sin2]$

$\displaystyle \frac{dy}{dx}(2x^2-cos2x-sin2) = -4xy$

$\displaystyle \frac{dy}{dx} = -\frac{4xy}{2x^2-cos2x-sin2}$

2. Should be product rule again on left-hand side, and I'm not at all sure what you're aiming at there on the right (should be chain rule). Balloons approaching...

Edit:

I see now that you were trying for the product rule but forgot the plus sign. Anyway...

Just in case it helps to see the formulae as well...

and

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3. $\displaystyle 2x^2y=sin2x$

$\displaystyle \frac{d}{dx}[2x^2y] = \frac{d}{dx}[sin2x]$

$\displaystyle 4xy+\frac{dy}{dx}[2x^2]=\frac{dy}{dx}[2cos2x]$

$\displaystyle \frac{dy}{dx}(2x^2-2cos2x) = 4xy$

$\displaystyle \frac{dy}{dx} = \frac{4xy}{2x^2-2cos2x}$

4. You still want to see a dy/dx on the right hand side? Not so...

LHS good.

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5. I keep it on both sides so that I later remember what it was, then I move everything to one side.

6. Yes but don't put it in if it doesn't belong... are you confusing dy/dx with d/dx? It's leading you down a wrong turn - your next line is way off

Edit:

Your factorising and so on are fine - subject the bottom row of my pic to the same processes and you should arrive at the correct derivative. Then have a think about why you were taking a different turn... probably the d/dx notation, which actually just means 'derivative (with respect to x) of...' (You don't need it anyway if you draw the picture!)

7. yeah, I think I'm confusing it with d/dx. Which line is way off? I though I had it right already

8. I get confused when doing this derrivative.

$\displaystyle \frac{d}{dx}[sin2x]$

explain please, I think you already did, but my brain is not absorbing it.

9. $\displaystyle \frac{d}{dx}\sin(2x)$ just means the derivative of sin(2x).

The chain rule says that the derivative of sin(anything) is cos(the same thing) times the derivative of the thing...

Try to read d/dx as just saying 'the derivative of' - or use a picture and do without it, because d/dx of anything is whatever's in the balloon below...

10. $\displaystyle 4xy+\frac{dy}{dx}[2x^2]=2cos2x$

$\displaystyle \frac{dy}{dx} = \frac{2cos2x-4xy}{2x^2}$

11. Yep

12. I know why I got confused when doing that part. It now shouldn't happen again.

Thanks alot