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

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you could solve for y then differentiate that way or you could do implicit if you wanted.
is this |dw:1359321531751:dw|
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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Other answers:

Yes thats right
looks like no one wants to help me here.
it goes something like |dw:1359321813888:dw|
so then solve for y'
double check what the derivative of tanx is too, i'm not 100% sure
ahan ok and I also have an answer for this that my professor gave which is: \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]
that's what you'll get once you rearrage and solve for y'
derivative of tanx is \[\sec ^{2}x\]
okay. now you know how to do it.
no still not sure
sorry Pinky - i got called away malcolm is explaining it well
can we go through till the answer
its ok cwrw238
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.
hmmmm but can you please also show me to rearrange it if its possible for you?
oops forgot the tanx in there but you just move the one term to the other side then divide by 2ytanx
that last drawing was completely wrong, i don't have time to write it out. it's really basic stuff you should know.
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
that's your full solution. don't come on here and expect people to add 1+1 for you.

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