Calculus problem....solved...for now.

arcath

Okie here is my problem.....

I am having to retake calculus 1 since my grade, which was a B, didn't transfer over.

And since this is my first math class in 7 years, along with the fact that this is a online class has only really stumped me thus far.

So I am having problems with implicit differentiation.

Working through the book, I got stumped at problem 12.

( (3(x^2)(y^2)) + (4(y^5)) + (8(x^2)(y^3)) + ((x)(y)) ) = 5

I just don't get how to solve this, and my TI-89 is off no help either because it cant solve Implicit Differentiation.

Anyone able to help me?

Iceman.USAF

implicit is just regular old differentiation right?

Have you tried isolating the x or y components yet? That would be my first move. Then you'll have a y=some equation format, and try and go from there.

Abels

Easy. Just remember to multiply parts containing y by y' (because y is a function of x -> chain rule), and you're golden. Once you're done, rearrange the equation to find an expression for y', and sub it back in.

TechBoy

Implicit differentiation is the same thing as regular differentiation. The difference being that say in a regular equation you're given to derive like y = x^3 - 2x, the X's and the Y's are separate. In implicit, the X's and Y's are all jumbled together and can't be separated.

However, that's no big deal. Simply treat X and Y as variables.

Things to note:

While the derivative of 2x is 2, and x^2 is 2x, the derivative of 2y is 2[dy/dx] and y^2 is 2y[dy/dx]. Basically, whenever you derive a y, act like you're deriving x, but also tack on a dy/dx.

Also, because you're treating both x and y as variables, deriving xy require the multiplication rule. The derivative of xy is y + x[dy/dx]

Lastly, I think after you go through the equation and derive everything, you're supposed to solve for dy/dx. So your answer should look like dy/dx = (lots o x's and y's)

Big Dookie

In case you just want the solution, here you go:

3(x^2)(y^2) + 4(y^5) + 8(x^2)(y^3) + xy = 5

Differentiate all terms, using the rules Techboy stated above:
3[(x^2)(2y)(y')+(y^2)(2x) ] + 4(5(y^4))(y') + 8[(x^2)(3(y^2))(y')+(y^3)(2x)] + [(x)(y')+(y)] = 0

Move all terms containting y' to one side and everything else to the other side, then factor out y'.
(y') [6(x^2)(y) + 20(y^4) + 24(x^2)(y^2) + x] = - [6(x^2)(y^2) + 16(x)(y^3) + y]

And finally, divide to solve for y'.
y' = - { [6(x^2)(y^2) + 16(x)(y^3) + y] / [6(x^2)(y) + 20(y^4) + 24(x^2)(y^2) + x] }

Note, I haven't done implicit differentiation in forever, so if I made a mistake, please forgive me.

arcath

Thanks much Dookie, the solution is nice, but I'm looking for the workthrough.

not having had a math class in so long, i am having to relearn it all. and seeing the step by step helps me out a lot.

thanks much.

b0bd0d

I feel your pain. I'm studying that implicit shit right now for a test tuesday. Um, this might help:
http://tutorial.math.lamar.edu/

Big Dookie

arcath wrote: »

Thanks much Dookie, the solution is nice, but I'm looking for the workthrough.

not having had a math class in so long, i am having to relearn it all. and seeing the step by step helps me out a lot.

thanks much.

No problem. However, that did have most of the work in it - I only skipped a couple of small algebra steps. Is there anything in particular you aren't understanding? The "Calculus" part of it is fairly easy - the same basic rules of differentiation apply, except that anywhere you see a "y", you just tack on a " y' ". For example:

(3x)(4y^2) = 8x

The left side of the equation is a product, so you use the product rule - f(x)g(x) = f(x)g'(x) + g(x)f'(x).

(3x)(8y) + (3)(4x^2) = 8

And now you multiply the derivative of the y-term by y'.

(3x)(8y)(y') + (3)(4x^2) = 8

Move any terms that don't contain y' to the right side of the equation, and simplify:

24xy(y') = 8 - 12x^2

And finally, divide so that you have the y' alone on the left side of the equation:

y' = (8 - 12x^2) / (24xy)

That's pretty much it. As you can see, other than the initial differentation and remembering to multiply y-derivatives by y', it's all pretty much algebra.