Pinky1234
Can anyone help me with this question? Y^2tan(x)=x. Find dy/dx.
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malcolm11235
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you could solve for y then differentiate that way or you could do implicit if you wanted.
cwrw238
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is this
|dw:1359321531751:dw|
Pinky1234
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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.
Pinky1234
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Yes thats right
Pinky1234
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looks like no one wants to help me here.
malcolm11235
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it goes something like |dw:1359321813888:dw|
malcolm11235
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so then solve for y'
malcolm11235
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double check what the derivative of tanx is too, i'm not 100% sure
Pinky1234
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ahan ok and I also have an answer for this that my professor gave which is: \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]
malcolm11235
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that's what you'll get once you rearrage and solve for y'
Pinky1234
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derivative of tanx is \[\sec ^{2}x\]
malcolm11235
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okay. now you know how to do it.
Pinky1234
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no still not sure
cwrw238
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sorry Pinky - i got called away
malcolm is explaining it well
Pinky1234
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can we go through till the answer
Pinky1234
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its ok cwrw238
malcolm11235
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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.
Pinky1234
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hmmmm but can you please also show me to rearrange it if its possible for you?
malcolm11235
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oops forgot the tanx in there
but you just move the one term to the other side then divide by 2ytanx
malcolm11235
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that last drawing was completely wrong, i don't have time to write it out. it's really basic stuff you should know.
malcolm11235
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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
Pinky1234
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ok
malcolm11235
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|dw:1359322857339:dw|
malcolm11235
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that's your full solution. don't come on here and expect people to add 1+1 for you.