1. Trouble finding basic derivative

Hi not sure if this answer is correct. Find derivative with respect to x:
(2x - 3y)^3 = x^2 - y

steps:
[(2x + 3y)^3 + y]^1/2 = x

1/2[(2x + 3y)^3 + y]^-1/2 * u'

u' = 3(2x + 3y)^2 * 5

1/2[(2x + 3y)^3 + y]^1/2 * 3(2x + 3y)^2 * 5

correct?

2. Re: Trouble finding basic derivative

Maybe I'm missing something but this looks like a problem in implicit differentiation to me (solving for y in terms of x might be possible, but what a chore). Multiply out the cubic then differentiate both sides by x. Then solve for y' (or dy/dx if you prefer that notation). It'll still be a mess, a rational expression with a couple of xy terms.

3. Re: Trouble finding basic derivative

the question states "find the derivative of the functions with respect to x"

4. Re: Trouble finding basic derivative

And what is the function of which you are finding the derivative? Is it not y(x)?

5. Re: Trouble finding basic derivative

you have as much information as I have unfortunately.

6. Re: Trouble finding basic derivative

Are you talking about implicit differentiation in class? I know this is a sticking point for many students.

When we first learn about differentiation, usually we represent the (dependent) function to be differentiated by some variable (y most often) and define it in terms of an (independent) variable (x most often), as

y=sin(x)

... but it's possible to differentiate a function without representing it that way. In the equation you gave, if x changes and the equation remains true, then y will change, ie, it will have a derivative with respect to x. That's true even y is not a function of x. In a circle like

x^2+y^2 = 1

there will be two values of y for each x (wherever the circle exists). At each point on the circle, y has a derivative with respect to x. For each x where the circle is defined, y will have two derivatives, one for each point where a vertical line cuts the circle. Is that clear?

If you plot y against x in your problem, you'll see that y isn't a function of x. It fails the vertical line test at some points. This means the derivative dy/dx will depend on the y coordinate as well as the x. If you're not clear on implicit differentiation, you can go back to your text, or I'm sure there are many materials on the web to help you.

One thing I noticed. You introduce a variable u, which is OK, but you don't explicitly tell us what u is. I gather that

u = (2x+3y)^3 + y

in which case

u' = 3*(2x + 3y)^2*(2+3y')+y'

Is that clear?

I can send you a plot of y against x if you can't generate one yourself.

7. Re: Trouble finding basic derivative

not too clear. I should probably plot the graph, but that will depend on time constraints. Have to go for the highest potential marks first - midterm on tuesday. This looks like a difficult question and faster progress is being made in other areas... more areas to cover.

8. Re: Trouble finding basic derivative

Originally Posted by togo
Hi not sure if this answer is correct. Find derivative with respect to x:
(2x - 3y)^3 = x^2 - y
Differentiating both sides with respect to x,
3(2x- 3y)^2(2- 3y')= 2x- y'

6(2x- 3y)^2- 9(2x-3y)y'= 2x- y'
6(2x- 3y)^2- 2x= [9(2x- 3y)- 1]y'

y'= (6(2x-3y)^2- 2x)/(9(2x- 3y)- 1]

You might want to multiply those out: (2x- 3y)^2= 4x^2- 12xy+ 9y^2 so 6(2x- 3y)^2- 2x= 24x^2- 12xy+ 9y^2- 2x and 9(2x- 3y)- 1= 18x- 27y- 1 .

steps:
[(2x + 3y)^3 + y]^1/2 = x

1/2[(2x + 3y)^3 + y]^-1/2 * u'

u' = 3(2x + 3y)^2 * 5

1/2[(2x + 3y)^3 + y]^1/2 * 3(2x + 3y)^2 * 5

correct?[/QUOTE]

9. Re: Trouble finding basic derivative

Originally Posted by togo
Hi not sure if this answer is correct. Find derivative with respect to x: (2x - 3y)^3 = x^2 - y

$(2x-3y)^3=x^2-y$

$3(2x-3y)^2 (2-3y')=2x-y'$

$6(2x-3y)^2-9(2x-3y)^2 y'=2x-y'$

$[1-9(2x-3y)^2] y'=2x-6x(2x-3y)^2$

Now solve for $y'$.

10. Re: Trouble finding basic derivative

The attached file contains a graph. Maybe it'll help you follow it. Plato's approach is quite correct.