Implicit differentiation

[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]

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

 

[karush]

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

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

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

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 \(\displaystyle x\). Use the product rule on \(\displaystyle 2xy\) (and the chain rule on \(\displaystyle y\)). Don't forget to post all of your work.

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

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 \(\displaystyle x\) in brackets there? Remove that and you have the correct result.