need help with the following derivitave

[sigma]

Find an equation of the tangent line to the curve \(\displaystyle \L\sin(xy)=\L\frac{\
\sqrt[3]{3}
\

}{2\pi }xy\)
at the point \(\displaystyle \L(\pi,\L\frac{1}{3})\)

So first, \(\displaystyle \L\frac{dy}{dx}\) needs to be calculated. The answer came out to be \(\displaystyle \L-\frac{1}{3\pi}\) which of course have no idea how they arrived at that answer. Tried doing this question with no luck. Hope somebody can help

 

[galactus]

Use implicit differentiation.

Are you familiar with it?. Believe it or not, using implicit differentiation this

ominous looking function breaks down to a very simple derivative.

 

[sigma]

Use implicit differentiation.

Are you familiar with it?. Believe it or not, using implicit differentiation this

ominous looking function breaks down to a very simple derivative.

Very familiar with it. Like I said, I tried doing this question with no luck. Just got some ugly answer. I tried differentiating the sin(xy) part then the other side. Never really done implicit differentiation with a function like this though, especially cube roots with a fraction which are fairly new to me with implicit differentiation. Wouldn't differentiating the cube root of 3 over 2 pi just be 0 because its constant? But I know the product rule applies here. Would the chain rule apply to sin(xy)?

 

[galactus]

Let's see what happens. Follow along.

Use the chain rule and the product rule:

\(\displaystyle \L{xcos(xy)}\frac{dy}{dx}+ycos(xy)+\frac{3^{\frac{1}{3}}}{2{\pi}}y+\frac{3^{\frac{1}{3}}}{2{\pi}}x\frac{dy}{dx}=0\)

\(\displaystyle \L{xcos(xy)}\frac{dy}{dx}+\frac{3^{\frac{1}{3}}}{2{\pi}}x\frac{dy}{dx}=-ycos(xy)-\frac{3^{\frac{1}{3}}}{2{\pi}}y\)

Divide through and factor:

\(\displaystyle \L\frac{-y(cos(xy)+\frac{3^{\frac{1}{3}}}{2{\pi}})}{x{(cos(xy)+\frac{3^{\frac{1}{3}}}{2{\pi}})\)

As you can see, doing the cancellation we are left with -y/x

Enter in \(\displaystyle \L\frac{\frac{{-}1}{3}}{{\pi}}=\frac{-\1}{3{\pi}}\)

 

[sigma]

Thanks allot. Thats almost what I did, well not really but on the right track. Heres what I did.

\(\displaystyle \
\L\cos (xy)(y)(x\frac{{dy}}{{dx}}) = \frac{{\sqrt[3]{3}}}{{2\pi }}y + \frac{{\sqrt[3]{3}}}{{2\pi }}x\frac{{dy}}{{dx}}
\\)

\(\displaystyle \
\L\cos (xy)(y)(x\frac{{dy}}{{dx}}) - \frac{{\sqrt[3]{3}}}{{2\pi }}x\frac{{dy}}{{dx}} = \frac{{\sqrt[3]{3}}}{{2\pi }}y
\\)

\(\displaystyle \
\L\frac{{dy}}{{dx}}(\cos (xy)(y)(x) - \frac{{\sqrt[3]{3}}}{{2\pi }}x)= \frac{{\sqrt[3]{3}}}{{2\pi }}y
\\)

\(\displaystyle \
\L\frac{{dy}}{{dx}} = \frac{{\frac{{\sqrt[3]{3}}}{{2\pi }}y}}{{(\cos (xy)(y)(x) - \frac{{\sqrt[3]{3}}}{{2\pi }}x}}
\\)

And then I stoped there because it didn't make any sense to me. I tried plugging in the values for x and y which created more mess and confusion unless you can do something with this! :shock:
\(\displaystyle \
\L\frac{{dy}}{{dx}} = \frac{{\frac{{\sqrt[3]{3}}}{{2\pi }}\frac{1}{3}}}{{(\cos (\frac{\pi }{3})(\frac{1}{3})(\pi ) - \frac{{\sqrt[3]{3}}}{{2\pi }}\pi }}
\\)

For what you did, where did the x and y in front of cos come from?

 

[galactus]

A combo of the chain rule, product rule, and factoring. is how that was

done.

\(\displaystyle \L\sin(xy)=cos(xy)(x\frac{dy}{dx}+y(1))=xcos(xy)\frac{dy}{dx}+ycos(xy)\)

Use the product rule on the other part, take the parts without dy/dx and

put them on the RHS; On the LHS, factor out dy/dx and divide through to

isolate dy/dx; Do any factoring that can be done on the RHS.

 

[sigma]

I see. So after you did the chain rule, you multiplied cos(xy) through. Do you have to always do that? I just learned implicit differentiation about a month ago and they always taught us not to multiply through after doing the chain rule unless your simplifying through which I guess in this case you are.

One other thing I'm missing here though. When you brought the RHS \(\displaystyle \L\frac{3^{\frac{1}{3}}}{2{\pi}}y+\frac{3^{\frac{1}{3}}}{2{\pi}}x\frac{dy}{dx}\) over to the LHS, how come its not negative? That is the same thing I got for the RHS then I brought it over to the LHS and made it negative so that's why I'm wondering.

 

[galactus]

Do you mean after I factored?. I factored out negative y. That means I had to make the negative a plus inside the parentheses.

 

[sigma]

No I meant right after you did the implicit differentiation (or before), it looks like you moved the RHS over to the LHS but didn't negate it (that is why is says equal to "0"). Just want to know why this was done.

 

[tkhunny]

Does it bother anyone that \(\displaystyle (\pi,\frac{1}{3})\) isn't ON the curve? Maybe I'm just missing something.

 

[galactus]

No I meant right after you did the implicit differentiation (or before), it looks like you moved the RHS over to the LHS but didn't negate it (that is why is says equal to "0"). Just want to know why this was done.

Sorry, that was a small booboo. All that does in change the signs inside the parentheses from positive to negative. The cancellations are the same.
You still wind up with -y/x.

Check out what tkh stated. I didn't check. I just performed the differentiation.