Can anyone help me with this question? Y^2tan(x)=x. Find dy/dx.

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- anonymous

Can anyone help me with this question? Y^2tan(x)=x. Find dy/dx.

- schrodinger

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- anonymous

you could solve for y then differentiate that way or you could do implicit if you wanted.

- cwrw238

is this
|dw:1359321531751:dw|

- anonymous

yeah we need to do implicit but can you go through with me because this is a new course for me and I am having really tough time with this. I have couple more questions then I ll try by myself.

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- anonymous

Yes thats right

- anonymous

looks like no one wants to help me here.

- anonymous

it goes something like |dw:1359321813888:dw|

- anonymous

so then solve for y'

- anonymous

double check what the derivative of tanx is too, i'm not 100% sure

- anonymous

ahan ok and I also have an answer for this that my professor gave which is: \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]

- anonymous

that's what you'll get once you rearrage and solve for y'

- anonymous

derivative of tanx is \[\sec ^{2}x\]

- anonymous

okay. now you know how to do it.

- anonymous

no still not sure

- cwrw238

sorry Pinky - i got called away
malcolm is explaining it well

- anonymous

can we go through till the answer

- anonymous

its ok cwrw238

- anonymous

i did go through to the answer. i just didn't rearrange it to solve for dy/dx= y', which is basic algebra, a prerequisite for your calculus course.

- anonymous

hmmmm but can you please also show me to rearrange it if its possible for you?

- anonymous

oops forgot the tanx in there
but you just move the one term to the other side then divide by 2ytanx

- anonymous

that last drawing was completely wrong, i don't have time to write it out. it's really basic stuff you should know.

- anonymous

that's not the answr there. thats me saying that y' = dy/dx in case you didn't know that
so in the equation, y' = dy/dx

- anonymous

ok

- anonymous

|dw:1359322857339:dw|

- anonymous

that's your full solution. don't come on here and expect people to add 1+1 for you.

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