Find the derivative of the function:

Attachment 20644

at the point (-1,2). Therefore need to find the equation of the tangent at that point but im having a hard time differentiating it first. Any help would be great

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- Jan 31st 2011, 05:49 AMHockey_Guy14Implicit Differentiation
Find the derivative of the function:

Attachment 20644

at the point (-1,2). Therefore need to find the equation of the tangent at that point but im having a hard time differentiating it first. Any help would be great - Jan 31st 2011, 05:59 AMBAdhi
$\displaystyle \frac{d(x^2y)}{dx}+\frac{d(y^2x)}{dx}=0$

use $\displaystyle \frac{d(uv)}{dx}=\frac{du}{dx}v+\frac{dv}{dx}u $for each part.

and use chain rule to get $\displaystyle \frac{dy^2}{dx}$

__Spoiler__: - Jan 31st 2011, 06:35 AMHockey_Guy14
Im sorry im still lost having problems with implicit differentiation cannot seem to figure out how to do it properly :(

- Jan 31st 2011, 06:59 AMHockey_Guy14
For example when you give 2xy + x^2dy/dx + 2ydy/dx x + y^2 = 0, what does the dy/dx mean? Does that mean differentiate further and if so how would I do that? would the x^2 become 2x and the 2yx become 2x or 2y?

- Jan 31st 2011, 06:59 AMBAdhi
please mention which part you are having problems with? is it at the begining or at the middle? because there is no point of just explaining every thing that both of us know

- Jan 31st 2011, 07:05 AMe^(i*pi)
dy/dx is a product of the chain rule and it becomes a term in it's own right. You then use algebra to isolate dy/dx.

2xy is neither. You need to use the product rule (which is in post 2) - Jan 31st 2011, 07:32 AMHockey_Guy14
having problems in the middle getting to the final answer. trying my best to figure out what you mean and how you are doing it and using the chain and product rules. If you cant help any further thats fine I appreciate the help you have given me so far.

- Jan 31st 2011, 07:42 AMBAdhi
Hey Hockey_Guy,

i apologize for posting the post #5 'cause i only saw your post #3.

keep practising chain and product rule and differentiating equations, hope you'll get it right :) - Jan 31st 2011, 07:50 AMHockey_Guy14
Thats no problem at all i understand Badhi, would you be able to give me the final answer however, and the equation of the tangent, if you have it, that way I can work through it and know what im trying to get and work with that? And then I can apply what im doing to other problems. If you can do that, that would be awesome would greatly appreciate it because I dont have a final final answer to sort of use as a guide to. Thanks!

- Jan 31st 2011, 07:56 AMBAdhi
all you got to do is substitute the x,y with the point (-1,2) to equation in the post #2 under the spoiler and isolate $\displaystyle \frac{dy}{dx}$ which will give the tangent of the curve at that point

- Jan 31st 2011, 08:13 AMHockey_Guy14
Alright thanks for your help :)