MHB Why Is Implicit Differentiation of This Equation So Tricky?

[karush]

$$6x-\sqrt{2xy}+xy^3 ={y}^{2}$$
$$6-?+3x{y}^{2}{y'}^{}+{y}^{3}=2y{y'}^{}$$

Got stumped on this one answer was complicated...

 

[Prove It]

$$6x-\sqrt{2xy}+xy^3 ={y}^{2}$$
$$6-?+3x{y}^{2}{y'}^{}+{y}^{3}=2y{y'}^{}$$

Got stumped on this one answer was complicated...

You're almost there. Write $\displaystyle \begin{align*} \sqrt{2\,x\,y} = \sqrt{2} \, x^{\frac{1}{2}} \, y ^{\frac{1}{2}} \end{align*}$, then you can use the product rule to differentiate it.

 

[karush]

$$u'v+vu'$$
$$\sqrt{2xy} \ '=\frac{\sqrt{2x}}{2\sqrt{y}}y'+\sqrt{2y}$$
Sorta?

 

[MarkFL]

I have moved this thread since our "Calculus" forum is a better fit.

 

[karush]

Oops, I thot it was in calculus??

 

[MarkFL]

Oops, I thot it was in calculus??

You overshot by one forum and went to "Business Mathematics" instead. :D

 

[Greg]

$$\dfrac{\mathrm d}{\mathrm dx}\sqrt{2xy}=\dfrac{\mathrm d}{\mathrm dx}(2xy)^{1/2}=\dfrac12(2xy)^{-1/2}\cdot\dfrac{\mathrm d}{\mathrm dx}(2xy)$$

 

[karush]

$$\dfrac{\mathrm d}{\mathrm dx}\sqrt{2xy}=\dfrac{\mathrm d}{\mathrm dx}(2xy)^{1/2}=\dfrac12(2xy)\cdot\dfrac{\mathrm d}{\mathrm dx}(2xy)$$

Is this the product rule?

 

[Greg]

It's the chain rule. Use the product rule to complete the differentiation.

 

[Prove It]

Or you could just follow my advice, thereby you can avoid the chain rule altogether...

 

[MarkFL]

Or you could just follow my advice, thereby you can avoid the chain rule altogether...

You would still need the chain rule...but I know what you mean. :D

 

[karush]

Is post #3 correct?

 

[Prove It]

You would still need the chain rule...but I know what you mean. :D

Only on the y term :P

 

[Greg]

My post (#7) contains an error. It should be

$$\dfrac{\mathrm d}{\mathrm dx}\sqrt{2xy}=\dfrac{\mathrm d}{\mathrm dx}(2xy)^{1/2}=\dfrac12(2xy)^{-1/2}\cdot\dfrac{\mathrm d}{\mathrm dx}(2xy)$$

Sorry about the confusion.

 

[karush]

$$
\displaystyle \dfrac{\mathrm d}{\mathrm dx}\sqrt{2xy}
=\dfrac{\mathrm d}{\mathrm dx}(2xy)^{1/2}
=\dfrac12(2xy)^{-1/2}\cdot\dfrac{\mathrm d}{\mathrm dx}(2xy)
$$

so could this be rewritten as
$$\left[\frac{\sqrt{2xy}}{2}\right]y'$$

 

[Greg]

No. You need to differentiate implicitly with respect to $$x$$. Use the product rule on $$2xy$$ (and the chain rule on $$y$$). Don't forget to post all of your work.

$$\dfrac12(2xy)^{-1/2}=\dfrac{1}{2\sqrt{2xy}}$$

It may very well be easier to use the method outlined by Prove It. Again, show all of your work.

 

[karush]

$$\d{}{x}\left(2xy\right)=2\left(xy'\left(x\right)+y\right)$$

 

[Greg]

Why do you have the $$x$$ in brackets there? Remove that and you have the correct result.