# A couple of simple derivative problems

• Mar 31st 2011, 06:30 PM
TenaciousE
A couple of simple derivative problems
It's been a while since I've done derivatives and I'm having problems with these two. I know they aren't horribly complicated, but I just can't seem to get them right

1) $tan(sqrt{(1-x)})$ Find Y'
2) $(x^2)(y^3) + 3y^2 = x - 4y$ Find Y''
• Mar 31st 2011, 06:34 PM
TheChaz
How do you know that you haven't gotten them right? Do you have the solutions? If so, please show your work and many of us would be glad to add our input.
• Mar 31st 2011, 06:56 PM
TenaciousE
Sorry I don't know how to use the text functions to make it look better >.< Anyway, here is what I have so far.
1) y = tan(sqrt(1-x))
y' = .5tan(1-x)^-.5 (sec(1-x))^2
y' = ((sec(1-x))^2)/2tan(1-x)^.5

2) I don't remember much at all about implicit differentiation.. I'm having trouble finding the first derivative
(x^2)(y^3) + 3y^2 = x - 4y
y'((3x^2)(y^2) + 6y + 4) = 1 - 2xy^3
y' = (1 - 2xy^3)/((3x^2)(y^2) + 6y + 4)
I'm not sure if either of these are actually right, but they don't look right. If the second one is right, the second derivative is going to be ridiculous
• Mar 31st 2011, 07:09 PM
TheChaz
Quote:

Originally Posted by TenaciousE
Sorry I don't know how to use the text functions to make it look better >.< Anyway, here is what I have so far.
1)
$y = tan(\sqrt{1-x})$
$y' = .5tan((1-x)^{-.5}) sec^2(1-x)$

$y' = \frac{(sec(1-x))^2)}{2tan(1-x)^{.5}}$
2) I don't remember much at all about implicit differentiation.. I'm having trouble finding the first derivative
$(x^2)(y^3) + 3y^2 = x - 4y
$

$y'((3x^2)(y^2) + 6y + 4) = 1 - 2xy^3$
$
y' = (1 - 2xy^3)/((3x^2)(y^2) + 6y + 4)$

I'm not sure if either of these are actually right, but they don't look right. If the second one is right, the second derivative is going to be ridiculous

(Many things clean-up nicely with tex tags. For exponents with more than one character, I used - and you should use! - { } )

I tried to make (1) look ok...
For the sake of my typing, let's call the square root "u".
We have y = tan(u)
Then y' = sec^2(u)*u'
And u' = -.5/u (don't forget to use chain rule TWICE to get the derivative of (1-x)).

2.
$(x^2)(y^3) + 3y^2 = x - 4y
$

$x^2(3y^2)y' + 2x(y^3) + 6y*y' = 1 -4y'$

$y'(3x^2y^2 + 6y + 4) = 1 - 2xy^3$

I think (trying to format and work it as I go...)! Then divide by the factor next to y'...