# Math Help - differential problem

1. ## differential problem

Differentiate the equation with respect to the variable and express in terms of . (Use D for ).

2. Well, do it.

The first term: $\frac{d}{dt}x^{4}\;=\;4\cdot x^{3} \cdot \frac{dx}{dt}$

You do the second term. You'll need the product rule.

3. Originally Posted by tbenne3
Differentiate the equation with respect to the variable and express in terms of . (Use D for ).
$x^4 + 3xy^2 = 8$

$\frac{d}{dt}(x^4 + 3xy^2) = \frac{d}{dt}(8)$

$\frac{d}{dt}(x^4) + \frac{d}{dt}(3xy^2) = 0$

$\frac{d}{dx}(x^4)\,\frac{dx}{dt} + 3x\,\frac{d}{dt}(y^2) + y^2\,\frac{d}{dt}(3x) = 0$

$4x^3\,\frac{dx}{dt} + 3x\,\frac{d}{dy}(y^2)\,\frac{dy}{dt} + y^2\,\frac{d}{dx}(3x)\,\frac{dx}{dt} = 0$

$4x^3\,\frac{dx}{dt} + 6xy\,\frac{dy}{dt} + 3y^2\,\frac{dx}{dt} = 0$

$(4x^3 + 3y^2)\,\frac{dx}{dt} + 6xy\,\frac{dy}{dt} = 0$

$6xy\,\frac{dy}{dt} = -(4x^3 + 3y^2)\,\frac{dx}{dt}$

$\frac{dy}{dt} = -\left(\frac{4x^3 + 3y^2}{6xy}\right)\frac{dx}{dt}$.

4. Whoops, a little off on the first term. Very clean, though.

5. Originally Posted by TKHunny
Whoops, a little off on the first term. Very clean, though.
Edited. Thanks.

6. says it is incorrect :\

7. Originally Posted by tbenne3
says it is incorrect :\
Check the edited version...

8. this is what i entered and it says incorrect: -((4x^3+3y^2)/(6xy))

9. Originally Posted by tbenne3
this is what i entered and it says incorrect: -((4x^3+3y^2)/(6xy))
That's because it's $-\left(\frac{4x^3 + 3y^2}{6xy}\right)D$.

If it's still wrong, go through my working and find my mistake - I've provided you a full solution so it shouldn't be too hard to check it's correct.

10. That's what "very clean" is all about. You can follow it and see where you may have wandered off. You can also gain great confidence that you have not wandered off. Sloppy can't do that.