I suppose this is implicit differentiation... The problem:

find yprime given $\displaystyle {e^{x^2}} + y^6 = {2x(cos(y))}$

so I then simplified it to

$\displaystyle {2e^{x^2}} + 6y^5 * (dy/dx)= 2{cos(y)}{2x(-sin(y))}(dy/dx)$

Is this okay? Are there any errors, or can I move forward and work the algebra?

2. Originally Posted by Awesomeo
I suppose this is implicit differentiation... The problem:

find yprime given $\displaystyle {e^{x^2}} + y^6 = {2x{cos{y}}}$

so I then simplified it to

$\displaystyle {2e^{x^2}} + 6y^5 * {y^,}= 2{cos{y}}{2x{-sin{y}}}$

Is this okay? Are there any errors, or can I move forward and work the algebra?

Sorry, you've got it all wrong =\

The derivative of $\displaystyle e^{x^2}$ is $\displaystyle 2x e^{x^2}$, by the chain rule.

The derivative of $\displaystyle 2x \cos(y)$, is, by the product rule, $\displaystyle 2 \cos(y) - 2x \sin(y) y'$

3. Originally Posted by Awesomeo
I suppose this is implicit differentiation... The problem:

find yprime given $\displaystyle {e^{x^2}} + y^6 = {2x{cos{y}}}$

so I then simplified it to

$\displaystyle {2e^{x^2}} + 6y^5 * (dy/dx)= 2{cos(y)}{2x(-sin(y))}(dy/dx)$

Is this okay? Are there any errors, or can I move forward and work the algebra?
Not quite.

$\displaystyle \frac{\text{d}}{\text{d}x}\left[e^{x^2}\right] = 2xe^{x^2}$

and you've done something really strange with your product rule. The RHS should read

$\displaystyle 2\cos y -2x\sin y\cdot \frac{\text{d}y}{\text{d}x}$

4. So I forgot the x in my $\displaystyle 2e^{x^2}$. I see that, but your RHS looks like mine I just can't figure out how to write in this math language.

2xe^x^2 +6y^5 = RHS stated below

5. Just in case a picture helps (and if not just ignore!) ...

... where

... is the chain rule, and...

... the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

On the right you have the chain rule wrapped inside the product rule - which you say you have ok. Actually I wouldn't have bothered posting except you missed out the chain in the second term - and I see now you didn't miss it at all first time round...

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: Gallery

Balloon Calculus Drawing with LaTeX and Asymptote!

6. I wish I could write fluently in this math language, but if you look at what I have so far, do I isolate the dy/dx's on the right side and bring everything else over to the other side? I'm pretty sure I keep gett dy/dx - dy/dx on the right when I simplify and move the non dy/dx's to the other side.

7. You should get

$\displaystyle \frac{\text{d}y}{\text{d}x}\left(6y^5-2x\sin y\right) = 2\cos y - 2xe^{x^2}$

8. so I'm seeing another one of my errors where I thought I had to differentiate the y inside the cos and the sin on the RHS. That's why I was getting dy/dx on both sides and when I would try to isolate them to a side, they would cancel each other.

Would I then simplify it so that the only thing on that one side of the equation is dy/dx then?

9. Yeah - just solve for $\displaystyle \frac{\text{d}y}{\text{d}x}$ as if it were any other algebra problem.