Find the equation of the tangent line of the curve

[Schaus]

Homework Statement

Find the equation of the tangent line to the curve $\ xy^2 + \frac 2 y = 4$ at the point (2,1).
Answer says $\ y-1 = -\frac 1 2(x-2)$
And with implicit differentiation I should have gotten $\frac {dy} {dx}= -\frac {y^2} {2xy-\frac {2} {y^2}}$

Homework Equations

$\ y-y_1 = m(x-x_1)$

The Attempt at a Solution

$\ y-1 = m(x-2)$

$\frac d {dx}(xy^2+\frac 2 y) =\frac d {dx}(4)$

$\frac d {dx}(xy^2)+\frac d {dx}(\frac 2 y) = 0$

$2xy+x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0$

$x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 2xy$

$\frac {dy} {dx}(x-\frac {2} {y^2})= 2xy$

$\frac {dy} {dx}= -\frac {2xy} {x-\frac {2} {y^2}}$
Subbing in my x and y values gives me a slope of 4.
If anyone can help show me what I did wrong, I would really appreciate it. Also this is the first time using LaTeX Primer so bare with me.

 

[FactChecker]

$2xy+x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0$

Check this. It looks wrong to me.

 

[Schaus]

$2xy+x⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0$ is this supposed to be
$x +2xy⋅\frac {dy} {dx}-\frac {2} {y^2}⋅ \frac {dy} {dx} = 0$?

 

[FactChecker]

Methodically applying the product rule to d/dx (x⋅y²) gives
d/dx (x⋅y²) = (d/dx x)⋅y² + x⋅(d/dx y²) = 1⋅y² + x⋅2⋅y⋅dy/dx = y² + 2xy⋅dy/dx
I don't see this in your calculations.

 

[Schaus]

So $\frac d {dx}(xy^2)+\frac d {dx}(\frac 2 y) = 0$ should turn into
$x\frac {d} {dx} (y^2)+ x⋅\frac d {dx} y^2- \frac 2 {y^2}(\frac {dy} {dx}) = 0$ because I don't see the derivative of $\frac 2 y$?

 

[Mark44]

So $\frac d {dx}(xy^2)+\frac d {dx}(\frac 2 y) = 0$ should turn into
$x\frac {d} {dx} (y^2)+ x⋅\frac d {dx} y^2- \frac 2 {y^2}(\frac {dy} {dx}) = 0$ because I don't see the derivative of $\frac 2 y$?

This part -- $x\frac {d} {dx} (y^2)+ x⋅\frac d {dx} y^2$ -- is wrong. You are not applying the product rule correctly. You have written the same thing twice.

 

[Schaus]

Ohhhh, I should have done this?
$\frac {d} {dx} (1⋅y^2)+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 y^2= 0$
then I would get $-\frac {y^2} {2xy-\frac 2 {y^2}}$ ?

 

[Mark44]

Ohhhh, I should have done this?
$\frac {d} {dx} (1⋅y^2)+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 {y^2}= 0$ Mod note: Edited the fraction in the last term on the left side.
then I would get $-\frac {y^2} {2xy-\frac 2 {y^2}}$ ?

No, that's not it either. Here's how the product rule works.
$\frac {d} {dx} (f(x) g(x)) = \frac {d} {dx}(f(x)) \cdot g(x) + f(x) \cdot \frac {d} {dx}(g(x)) = f'(x)g(x) + f(x)g'(x)$
Let's start from the beginning.
$\frac {d} {dx} (xy^2) = (\frac {d} {dx} x) \cdot y^2 + x \frac {d} {dx}(y^2)$
Can you continue? You'll need to use the chain rule to evaluate the derivative on the right.

 

[Ray Vickson]

Ohhhh, I should have done this?
$\frac {d} {dx} (1⋅y^2)+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 y^2= 0$
then I would get $-\frac {y^2} {2xy-\frac 2 {y^2}}$ ?

No, that is even worse.

Go back to the beginning, and use the product rule carefully, doing it one step at a time. Check your work as you go along.

 

[Schaus]

$\frac {d} {dx} x⋅y^2+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 {y^2}= 0$
then I would get $-\frac {y^2} {2xy-\frac 2 {y^2}}$
I thought that I did the product rule.
f(x) = x
f'(x) = 1
g(x) = y²
g'(x) = 2y
I just did the product rule as f'g⋅g'f.
After the product rule I add the $\frac 2 y$ but the derivative of this turns it into $-\frac 2 {y^2}$? I can't seem to see what I've done wrong.

 

[Ray Vickson]

$\frac {d} {dx} x⋅y^2+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 {y^2}= 0$
then I would get $-\frac {y^2} {2xy-\frac 2 {y^2}}$
I thought that I did the product rule.
f(x) = x
f'(x) = 1
g(x) = y²
g'(x) = 2y
I just did the product rule as f'g⋅g'f.
After the product rule I add the $\frac 2 y$ but the derivative of this turns it into $-\frac 2 {y^2}$? I can't seem to see what I've done wrong.

It does not turn into $-2/y^2$, it turns into $-2 y'/y^2$.

 

[FactChecker]

$\frac {d} {dx} x⋅y^2+\frac {dy} {dx}2y⋅x -\frac {dy} {dx} \frac 2 {y^2}= 0$
then I would get $-\frac {y^2} {2xy-\frac 2 {y^2}}$
I thought that I did the product rule.
f(x) = x
f'(x) = 1
g(x) = y²
g'(x) = 2y

No. Don't forget that a change of x forces a change of y because of the initial equation. The chain rule gives g'(x) = 2y $\frac {dy} {dx}$

I just did the product rule as f'g⋅g'f.
After the product rule I add the $\frac 2 y$ but the derivative of this turns it into $-\frac 2 {y^2}$?

Same problem here. Use the chain rule and include $\frac {dy} {dx}$.

 

[Schaus]

I think I'll just have to watch more videos on implicit differentiation because I'm lost.

 

[Schaus]

Ok, let me see if I've done this properly. The product rule - $(x\frac{dy}{dx}⋅2xy)+(1⋅y^2)$. Now for the $\frac 2 y$ do I use the quotient rule?

 

[Ray Vickson]

Ok, let me see if I've done this properly. The product rule - $(x\frac{dy}{dx}⋅2xy)+(1⋅y^2)$. Now for the $\frac 2 y$ do I use the quotient rule?

You could use the quotient rule, but it is a whole lot easier just to use the chain rule.

 

[Schaus]

Ok so the chain rule is $f'((g(x))⋅g'(x)$, correct? I'm a little confused as to what I am subbing in where.

 

[Mark44]

Ok so the chain rule is $f'((g(x))⋅g'(x)$, correct? I'm a little confused as to what I am subbing in where.

Here's the chain rule for the problem you're having, using Leibniz notation.
$\frac d{dx}(y^2) = \frac d {dy}(y^2) \cdot \frac{dy}{dx}$
Can you continue?

 

[FactChecker]

Ok so the chain rule is $f'((g(x))⋅g'(x)$, correct? I'm a little confused as to what I am subbing in where.

You can use the derivative of a power and the chain rule to calculate 2/y.

2/y is really 2⋅y^-1 and y is really a function of x, so it is 2⋅y(x)^-1
Therefore $\frac {d} {dx}$(2/y)
= 2⋅$\frac {d} {dx}$(y(x)^-1)
= 2⋅(-1)y(x)^-2⋅y'(x)
= (-2/y²)⋅$\frac {dy} {dx}$

 

[cristo]

Ok so the chain rule is $f'((g(x))⋅g'(x)$, correct? I'm a little confused as to what I am subbing in where.

It helps if you try to understand why you are doing what you are doing.

So, to differentiate $\frac{2}{y}$ with respect to x, you want to first differentiate with respect to y: $\frac{d}{dy}\Big(\frac{2}{y}\Big)$ and then multiply by dy/dx. So you have $\frac{d}{dx}\Big(\frac{2}{y}\Big)=\frac{d}{dy}\Big(\frac{2}{y}\Big)\cdot\frac{dy}{dx}$

 

[Schaus]

Ohhh ok so it should look something like this?

$\frac {d}{dx}(xy^2+\frac 2 y )= \frac d {dx}(4)$
$\frac {d}{dx}(xy^2)+\frac d{dx}\frac 2 y=0$
$(x\frac d{dx}⋅2xy)+(1⋅y^2)+(\frac d{dx}\frac 2 y )= 0$
$2xy⋅\frac {dy}{dx}+y^2-\frac 2 {y^2}⋅\frac {dy}{dx}=0$
$\frac {dy}{dx}(2xy- \frac 2 {y^2}) = -y^2$
$\frac {dy}{dx} = - \frac {y^2} {2xy- \frac 2 {y^2}}$
Have I done this correctly?

 

[Mark44]

Ohhh ok so it should look something like this?

$\frac {d}{dx}(xy^2+\frac 2 y )= \frac d {dx}(4)$
$\frac {d}{dx}(xy^2)+\frac d{dx}\frac 2 y=0$
$(x\frac d{dx}⋅2xy)+(1⋅y^2)+(\frac d{dx}\frac 2 y )= 0$
$2xy⋅\frac {dy}{dx}+y^2-\frac 2 {y^2}⋅\frac {dy}{dx}=0$
$\frac {dy}{dx}(2xy- \frac 2 {y^2}) = -y^2$
$\frac {dy}{dx} = - \frac {y^2} {2xy- \frac 2 {y^2}}$
Have I done this correctly?

No. The first two lines are correct, but things fall apart in the third line.
In differentiating $xy^2$ you are have a mistake here:

$(x\frac d{dx}⋅2xy)$

For this part of the product rule you should have $x\frac d {dx}(y^2)$. You'll need to use the chain rule as I described in a previous post to finish this part.

Also, do not write $\frac d{dx}⋅2xy$. You are not multiplying $\frac d {dx}$ and 2xy, which is what the dot (⋅) signifies.
In the line below the one I'm talking about, it looks like you corrected your mistake.

 

[FactChecker]

I would say that you need to work on mastering the basic rules of derivatives some more (product, chain, quotient). Almost all problems are solved using a combination of those. Until you get them mastered, there is little chance of getting more complicated combinations correct.

 

[Schaus]

Yes, I've been trying to master the basics. I'm hoping that khan academy can help with that.

 

[Ray Vickson]

Yes, I've been trying to master the basics. I'm hoping that khan academy can help with that.

Why not purchase an actual BOOK on the subject? You can learn a lot from (for example) the Schaum's Outline series. Also: there are lots of free calculus textbooks available on-line, as well as tutorials and course notes, etc.

 

[Schaus]

Why not purchase an actual BOOK on the subject? You can learn a lot from (for example) the Schaum's Outline series. Also: there are lots of free calculus textbooks available on-line, as well as tutorials and course notes, etc.

I did not know of any books available. The course I'm doing is not very good. Thanks for the info I'm more than willing to purchase a book if it'll help.

 

[FactChecker]

I am also a big fan of the Schaum's Outline series. In post #22, some basic rules I forgot to mention are the specific derivatives of xⁿ, e^x, ln(x), and the trig functions. You should memorize those.

 

[cristo]

I did not know of any books available. The course I'm doing is not very good. Thanks for the info I'm more than willing to purchase a book if it'll help.

What's the course text you are using?

 

[Schaus]

I'm doing an online course called "Studyforge" or at least created by them. The course lacks detailed solutions for many of their problems which is hard for me because I learn more easily from seeing something done. I ordered https://www.amazon.com/gp/f.html?C=...LWQVEWA&ref_=pe_386430_126088100_TE_item&tag= and I'm hoping it is as good as the reviews say.

 

[FactChecker]

I'm doing an online course called "Studyforge" or at least created by them. The course lacks detailed solutions for many of their problems which is hard for me because I learn more easily from seeing something done. I ordered https://www.amazon.com/gp/f.html?C=...LWQVEWA&ref_=pe_386430_126088100_TE_item&tag= and I'm hoping it is as good as the reviews say.

It looks like this has only problems and solutions. That may be fine for your needs. Schaum's Outline of Calculus has fewer problems and solutions (still over 1,000) along with some explanation and video access.

 

[Schaus]

So I got the wrong one? Crap

 

[Schaus]

Alright, well I just got both. Can't hurt to have more.

 

[FactChecker]

So I got the wrong one? Crap

I don't know.

 

[Mark44]

I learn more easily from seeing something done.

Certainly it's easier to understand a problem by seeing a solution to it, but I don't necessarily believe that you're actually learning much of anything. A good textbook will have plenty of examples, but the learning comes when you actually do the work for a problem. Part of the learning process is in making false starts and going along branches that don't lead anywhere. This thread is (I hope) a good example of what I'm talking about.