finding y' on two equations

[Dantes]

We are learning logs and expos in calc (everyone's favorite topic) and I got stuck on two problems.

First one is y = x^-x^2 and we have to find y'.

On this one I can't figure out how to start it. I know we are suppose to show our work and then get help here, but I just can't figure out how to pull y's out of that equation and then do it implicitly. I'd say chain rule somehow. but first do x^-2x and somehow do that.

Second one is y = (3x^2+2y^2) and we have to find y'.

on this one I think you do it so you get 1 / x * (3x^2 + 2y^2) * 6x + 4y' but then i get stuck, cause my algebra sucks.

thanks for any advice and a happy turkey day to all.

martin

[ehild]

First one is y = x^-x^2 and we have to find y'.



Use

x=e^{ln(x)} \mbox { so } y=x^{-x^2}=e^{-ln(x)x^2}

and then apply the chain rule.



Second one is y = (3x^2+2y^2) and we have to find y'.

on this one I think you do it so you get 1 / x * (3x^2 + 2y^2) * 6x + 4y' but then i get stuck, cause my algebra sucks.




You have to express y' by x and y. Do not overcomplicate things.

y'=6x+4yy'\rightarrow y'(1-4y)=6x\rightarrow y' = \frac{6x}{1-4y}



ehild

[Tide]

There are a couple of ways of approaching the first one including direct application of the chain rule. You might simplify matters somewhat by taking the logarithm of both sides first and then differentiating.

\ln y = -x^2 \ln x

so

\frac {y'}{y} = -2x \ln x - x

from which you can find y'.

For the second, just differentiate both sides

y' = 6x + 4yy'

which will let you find y'.