Beginner implicit differentiation

[Kawakaze]

Please go easy on me, 2 days ago I didnt even know what implicit differentiation was.

Homework Statement

If x tan y − y tan x = 1, use implicit differentiation to determine dy/dx, expressing your answer in the form

dy/dx = f(x, y),

The Attempt at a Solution

Differentiate first with respect to x, and then y as a function of x

1.sec²(y) + x.sec²(dy/dx) - y.sec²(1) + dy/dx.sec²(x) = 0

If correct, it is still meaningless to me. What exactly am I trying to achieve here?

 

[Char. Limit]

Please go easy on me, 2 days ago I didnt even know what implicit differentiation was.

Homework Statement

If x tan y − y tan x = 1, use implicit differentiation to determine dy/dx, expressing your answer in the form

dy/dx = f(x, y),

The Attempt at a Solution

Differentiate first with respect to x, and then y as a function of x

1.sec²(y) + x.sec²(dy/dx) - y.sec²(1) + dy/dx.sec²(x) = 0

If correct, it is still meaningless to me. What exactly am I trying to achieve here?

I think you're using the product rule incorrectly. In particular, the first term should result in tan(y) + x sec^2(y) y'.

 

[chiro]

Please go easy on me, 2 days ago I didnt even know what implicit differentiation was.

Homework Statement

If x tan y − y tan x = 1, use implicit differentiation to determine dy/dx, expressing your answer in the form

dy/dx = f(x, y),

The Attempt at a Solution

Differentiate first with respect to x, and then y as a function of x

1.sec²(y) + x.sec²(dy/dx) - y.sec²(1) + dy/dx.sec²(x) = 0

If correct, it is still meaningless to me. What exactly am I trying to achieve here?

The point of implicit differentiation is to use the chain rule to come up with the correct derivative.

Note that D(uv) = Du . v + Dv . u

With your term x tan(y)

D(uv) = d/dx(x) . tany + d/dx(tany) . x
= tany + sec^2(y) . x . dy/dx

Check your results with the other term.

 

[Kawakaze]

Hi Chiro, thanks for the quick reply. Where do you get the u and v from? From what I've seen in my books, we would use this to break up say sin(x^3). Then u would equal x^3, these examples I can follow.

Must i differentiate the equation with respect to x, and then add this to the same equation differentiated to y? My textbook would imply I multiply the terms.

Its written exactly like this in my textbook, and is called the composite rule, I think this is the chain rule.

\stackrel{dy}{dx} = \stackrel{dy}{du}\stackrel{du}{dx}

Is there differences in terminology, this forum is based in america, and i am based in europe. I am sure the procedures are the same, but little differences in terminology can sure put a spanner in the works

 

[Kawakaze]

Wait a sec, I missed char limits post

1.tan(y) + x.sec²(y) \stackrel{dy}{dx} - y.sec²(1) + y\stackrel{dy}{dx}.tan(x)

Maybe its because its 3am, I am having a hard time, this makes sense to me, kind of. But the result from wolfram alpha is completely different

http://www.wolframalpha.com/input/?i=differentiate+x+tan+y+−+y+tan+x+=+1

 

[chiro]

Hi Chiro, thanks for the quick reply. Where do you get the u and v from? From what I've seen in my books, we would use this to break up say sin(x^3). Then u would equal x^3, these examples I can follow.

Must i differentiate the equation with respect to x, and then add this to the same equation differentiated to y? My textbook would imply I multiply the terms.

Its written exactly like this in my textbook, and is called the composite rule, I think this is the chain rule.

\stackrel{dy}{dx} = \stackrel{dy}{du}\stackrel{du}{dx}

Is there differences in terminology, this forum is based in america, and i am based in europe. I am sure the procedures are the same, but little differences in terminology can sure put a spanner in the works

Thats the composition rule.

d/dx f(g(x)) = g'(x) . f'(g(x))

The rule I stated above is the product rule where u and v are both functions of the same variable (example u(x) and v(x))

 

[Mark44]

Thats the composition rule.

More commonly called the chain rule...

d/dx f(g(x)) = g'(x) . f'(g(x))

The rule I stated above is the product rule where u and v are both functions of the same variable (example u(x) and v(x))

 

[SammyS]

Wait a sec, I missed char limits post

1.tan(y) + x.sec²(y) \stackrel{dy}{dx} - y.sec²(1) + y\stackrel{dy}{dx}.tan(x)

Maybe its because its 3am, I am having a hard time, this makes sense to me, kind of. But the result from wolfram alpha is completely different

http://www.wolframalpha.com/input/?i=differentiate+x+tan+y+−+y+tan+x+=+1

\frac{d}{dx}\tan(x)=sec^2(x)\,\cdot\,(\frac{dx}{dx})=sec^2(x)\,\cdot\,(1)

Why to you have y\cdot\frac{dy}{dx} in the last term?

 

[Kawakaze]

Why to you have y\cdot\frac{dy}{dx} in the last term?

Thats a mistake =) so the final answer is

f(x.y) = (tan(y) + x.sec²(y)\stackrel{dy}{dx})-(y.sec²(x) + \stackrel{dy}{dx}tan(x))

?

 

[Mark44]

Forget the f(x, y).
You started with x tan(y) − y tan(x) = 1.
Differentiate both sides of this equation to get
tan(y) + xsec²(y) dy/dx - dy/dx * tan(x) - ysec²(x) = 0

EDIT: Fixed a typo in the line above.

Now group the terms with dy/dx on one side and the other terms on the right side, and solve algebraically for dy/dx.

 

[SammyS]

Thats a mistake =) so the final answer is

f(x.y) = (tan(y) + x.sec²(y)\stackrel{dy}{dx})-(y.sec²(x) + \stackrel{dy}{dx}tan(x))

?

Your derivative looks good, but this is not the f(x,y) you are looking for.

Taking the derivative w.r.t. x of both sides of your original equation gives:

\left(\tan(y)+x\cdot\sec^2(y)\,\cdot\,\frac{dy}{dx}\right)-\left(y\cdot\sec^2(x) + \frac{dy}{dx}\,\tan(x)\right)=0

Solve this for dy/dx. The result is f(x,y), a function of x and y.

 

[Kawakaze]

Heya Mark, Sammy. Thanks for the help I think I got it. If my boss knew how much time I spent figuring this out instead of working . .. =)

tan(y) + x.sec²(y)\stackrel{dy}{dx} - y.sec²(x) - \stackrel{dy}{dx}tan(x) = 0

Which simplifies to

\stackrel{y.sec²(x) - tan(y)}{x.sec²(y) - tan(x)} = 0

So the dy/dx terms cancel each other out, to leave a 0. This then qualifies as a legitimate f'(x, y) expression?

 

[Mark44]

Fixed the first of your LaTeX things. I can't figure out what you're trying to say in the second one, so I can't fix it.

Instead of stackrel, use \frac{}{}.

Heya Mark, Sammy. Thanks for the help I think I got it. If my boss knew how much time I spent figuring this out instead of working . .. =)

tan(y) + x.sec^2(y)\frac{dy}{dx} - y.sec^2(x) - \frac{dy}{dx}tan(x) = 0

Which simplifies to

\stackrel{y.sec²(x) - tan(y)}{x.sec²(y) - tan(x)} = 0

So the dy/dx terms cancel each other out, to leave a 0. This then qualifies as a legitimate f'(x, y) expression?

No, the dy/dx terms don't cancel out. You should be able to solve for dy/dx algebraically.

 

[Kawakaze]

Thats weird, the first one displays ok on my screen, the second one is messy. here we go. But if the dy/dx don't cancel, then I am confused again as that would leave

\frac{y.sec^2(x) - tan(y)}{x.sec^2(y) - tan(x)} = \frac{dy}{dx} - \frac{dy}{dx}

*edit, I am really starting to hate latex, everyone else makes it look so easy. :D

 

[Char. Limit]

Thats weird, the first one displays ok on my screen, the second one is messy. here we go. But if the dy/dx don't cancel, then I am confused again as that would leave

\frac{y.sec²(x) - tan(y)}{x.sec²(y) - tan(x)} = \frac{dy}{dx} - \frac{dy}{dx}

*edit, I am really starting to hate latex, everyone else makes it look so easy. :D

Well, you need to use the tex brackets. You're not using them here.

You're not using the distributive property correctly. What you're doing is:

\frac{ac-ab}{c-b} = a-a

The correct way is

\frac{ac-ab}{c-b} = a

When you have all the dy/dx terms on one side, factor out the dy/dx and you should have dy/dx * (a bunch of stuff). Then just divide by the bunch.

 

[Kawakaze]

When you have all the dy/dx terms on one side, factor out the dy/dx and you should have dy/dx * (a bunch of stuff). Then just divide by the bunch.

Do you mean this?

\frac{dy}{dx}(x.sec^2(y) - y.sec^2(x)) - tan(y) + tan(x) = 0

I don't see what to do next here at all

 

[Mark44]

This is from post #10

Forget the f(x, y).
You started with x tan(y) − y tan(x) = 1.
Differentiate both sides of this equation to get
tan(y) + xsec²(y) dy/dx - dy/dx * tan(x) - ysec²(x) = 0
EDIT: Fixed typo in line above.

Now group the terms with dy/dx on one side and the other terms on the right side, and solve algebraically for dy/dx.

 

[Kawakaze]

This is from post #10

I see that now, =) but back at post #10 I didnt get it, also where did the tan come from before the equals sign.

 

[Kawakaze]

This is from post #10

\frac{dy}{dx}(x.sec^2(y) - tan(x)) -y.sec^2(x) + tan(y) = 0

\frac{dy}{dx}(x.sec^2(y) - tan(x)) = y.sec^2(x) - tan(y)

\frac{dy}{dx} = \frac{y.sec^2(x) - tan(y)}{x.sec^2(y) - tan(x)}

I think it just clicked =)

If it did, thanks guys, really you are saints!

 

[Mark44]

I see that now, =) but back at post #10 I didnt get it, also where did the tan come from before the equals sign.

It's a typo that shouldn't be there. I'll fix that.

 

[Char. Limit]

\frac{dy}{dx}(x.sec^2(y) - tan(x)) -y.sec^2(x) + tan(y) = 0

\frac{dy}{dx}(x.sec^2(y) - tan(x)) = y.sec^2(x) - tan(y)

\frac{dy}{dx} = \frac{y.sec^2(x) - tan(y)}{x.sec^2(y) - tan(x)}

I think it just clicked =)

If it did, thanks guys, really you are saints!

Yes, that is correct. Nice job!

 

[Kawakaze]

Thanks again guys! You saved my bacon.

 

[Char. Limit]

Thanks again guys! You saved my bacon.

It's no problem at all. And thank you for contributing!

 

[SammyS]

Hey Kawakaze,

When you use LaTeX on this site: After clicking on "Preview Post", you need to have your browser Re-Post (or Re-fresh) your screen - if you want to see your changes for the LaTeX stuff.

It was good to work with you.

Mark44 and Char. Limit post really solid stuff.

 

[Char. Limit]

Hey Kawakaze,

When you use LaTeX on this site: After clicking on "Preview Post", you need to have your browser Re-Post (or Re-fresh) your screen - if you want to see your changes for the LaTeX stuff.

It was good to work with you.

Mark44 and Char. Limit post really solid stuff.

Why thank you. It's always good to get compliments. I've seen some of your work, and it tends to be solid as well.

 

[Mark44]

Thank you from me as well. It makes me feel warm and fuzzy to hear such nice compliments. Thanks, Char. Limit and SammyS and Kawakaze!