1. ## More differentiation problems

I solved a few more differentiation problems, but got stuck on the following questions:

Find an expression in tems of x and y for dy/dx, given that:

(x-y)^4=x+y+5

and

((xy)^0.5)+x+y^2=0

2. The Captain got it right, I goofed!

-Dan

3. Originally Posted by Showcase_22
I solved a few more differentiation problems, but got stuck on the following questions:

Find an expression in tems of x and y for dy/dx, given that:

(x-y)^4=x+y+5
Implicit differentiation should be your watchword.

$
\frac{d}{dx}(x-y)^4=\frac{d}{dx}(x+y+5)
$

so:

$
4(x-y)^3 \left(1-\frac{dy}{dx}\right)=1+\frac{dy}{dx}
$
.

Now rearrange into the required form.

RonL

4. Originally Posted by Showcase_22
$\sqrt{xy}+x+y^2=0$
$\left ( \frac{1}{2} \frac{1}{\sqrt{xy}} \right ) \cdot \left ( y + x \frac{dy}{dx} \right ) + 1 + 2y \frac{dy}{dx} = 0$
where the first set of parenthesis is the $\frac{df}{dg}$ and the second set is the $\frac{dg}{dx}$.

All that's left is to solve for $\frac{dy}{dx}$. I leave it to you. Please feel free to post if you have any difficulties with it.

-Dan

5. Originally Posted by topsquark
Again, this is a problem in implicit differentiation.

Recall the chain rule:
$\frac{d}{dx}f(g(x)) = \frac{df}{dg} \frac{dg}{dx}$

So the LHS of your problem becomes, upon differentiation:
$4(x - y)^3 \cdot - \frac{dy}{dx}$

?

RonL

6. Originally Posted by CaptainBlack
?

RonL
I saw what I did (or didn't do) when I caught your post. Ah well, it's been a morning already!

-Dan