Question

implicit differentiation
tan(xy)=x+y

Answer 1

Hey Tyler c:
\[\Large\rm \tan(xy)=x+y\]So where we gettin stuck at?

Answer 2

im trying to find the slope of the line tangent at (0,0)

Answer 3

but i dont know where to begin exactly

Answer 4

We're going to have a derivative function that depends on BOTH x and y.\[\Large\rm y'(x,y)\]We'll evaluate that derivative function at (0,0) and this will give us the slope, \(\Large\rm m\), of the line tangent to the curve at that point.
\(\Large\rm y'(0,0)=m\)

Answer 5

So we need to take a derivative, then to try solve for y'.

Answer 6

so we do
tan(xy)-x=y?

Answer 7

the find y prime?

Answer 8

No, you don't NEED TO move the x.
We're going to also get a y' popping out of the tangent, so it won't be as nice and neat as you would like.

Answer 9

\[\Large\rm \frac{d}{dx}\tan(xy)\]So what would be your first instinct for this?
Do you remember the derivative of tangent?

Answer 10

sec^2(u)

u=xy?

Answer 11

Ok good.
Turns out we need to also apply our chain rule though:\[\Large\rm \frac{d}{dx}\tan(xy)=\sec^2(xy)\cdot(xy)'\]

Answer 12

We have a `product` of x and y,
how should we deal with that? :O

Answer 13

thats where im getting stuck

Answer 14

Notice the word I used... `product`.
What rule lets us take the derivative of a `product`?
Comeonnnn you know this :3

Answer 15

ok, so x'y+y'x

Answer 16

\[\Large\rm (xy)'=(x)'y+x(y)'\]Ok good.
So start by taking the derivative of x with respect to x.
x' = ?

Answer 17

1

Answer 18

Good, that's just your normal power rule.\[\Large\rm (xy)'=(1)y+x(y)'\]

Answer 19

y+x?

Answer 20

The derivative of y, with respect to x... this is what we're calling our dy/dx.
So we have,\[\Large\rm \frac{d}{dx}\tan(xy)=\sec^2(xy)\cdot(y+xy')\]

Answer 21

We don't get 1 when we take the derivative of y.

Answer 22

That's the part that's confusing? :o

Answer 23

ya

Answer 24

Lemme just try to give an example real quick.
Technically, we're ALWAYS applying the chain rule, but when your variable matches the derivative operator, it doesn't give us anything significant.
\[\Large\rm \frac{d}{dx}(x)^2=2(x)^1\frac{d}{dx}x\]Here I'm applying the power rule,
then I'm applying the chain rule, multiplying by the derivative of the inner function.
We could write it like this:
\(\Large\rm =2(x)^1\dfrac{dx}{dx}\)
But this dx/dx term has no significance.
It basically is saying, "as x changes an immeasurably small amount, how much does x change?"

\(\Large\rm =2x\)

Answer 25

But with y, this extra chain is really important :O\[\Large\rm \frac{d}{dx}(y)^2=2(y)^1\frac{d}{dx}y\]\[\Large\rm =2y\frac{dy}{dx}\]

Answer 26

I like to just remember it this way (even though this isn't very rigorous),
Whenever you take the derivative of something and it contains a variable that doesn't match your derivative operator, you're getting a prime that pops out.

So whenever we take the derivative of \(\Large\rm y\) with respect to any other variable besides y, we're going to get a \(\Large\rm y'\) popping out.

This will become very important later in the semester when you deal with related rates.

Answer 27

Too confusing still? :c

Answer 28

you still get 2x and 2y for both though?

Answer 29

No :)
You get 2x for the first
and you get 2y y' for the other one.

Answer 30

We COULD write 2x x' for the first one... but again, the derivative of x with respect to x is just 1.

Answer 31

So for the right side of our equation we get,\[\Large\rm (x+y)'=x'+y'=1+y'\]The y' doesn't simplify in the same way the x' does.

Answer 32

so the derivative of y is just y'

Answer 33

derivative of x is 1

Answer 34

(y+y')sec^2(xy)=y'

Answer 35

?

Answer 36

Yes, since our derivative is being taken with respect to x.

Here's another couple of examples to maybe help nail the point.
\(\Large\rm (x^3)'=3x^2\)
\(\Large\rm (y^3)'=3y^2y'\)

Answer 37

What happened to your x derivatives? They disappeared :O

Answer 38

y+y'*sec^2(xy)=1+y'

Answer 39

On the left side we got:\[\Large\rm \sec^2(xy)\cdot(y+xy')\]And on the right side we got:\[\Large\rm 1+y'\]

Answer 40

\[\Large\rm \sec^2(xy)\cdot(y+xy')=1+y'\]To solve for y' from here is a little tricky.
Requires a few algebra steps.

Answer 41

subtract and divide everything, press ctrl alt delete, etc etc

Answer 42

We'll distribute the sec^2 to each term in the brackets.\[\Large\rm y \sec^2(xy) + x \sec^2(xy) y'=1+y'\](We usually write derivative terms on the right like this, hopefully that's not too confusing.

Answer 43

Yah, some of that stuff :3

Answer 44

Then get your derivative terms on the same side, using subtraction.
And the other stuff to the other side.\[\Large\rm y \sec^2(xy)-1=y'-x \sec^2(xy)y'\]

Answer 45

Factor out your y',\[\Large\rm y \sec^2(xy)-1=y'\left[1-x \sec^2(xy)\right]\]

Answer 46

And then to solve for y', simply divide, yah?

Answer 47

That gives you your derivative function y'.
Then they want to know what the value of y' is when x=0, y=0.

Answer 48

And now that I think about it.....
You could have plugged in x=0, y=0 immediately after taking your derivative.
Things may have simplified on their own.

Answer 49

-1?

Answer 50

Mmm yah that sounds right! :)\[\Large\rm y'(0,0)=-1\]

Answer 51

dang..

Answer 52

Bro bro bro.
Chain rule is, by far, the hardest of the derivative rules to master.
Really spend some time with it! :O

This y' shenanigans will getcha!

Answer 53

alright

Answer 54

Here is what the graph looks like, just in case that helps.
https://www.desmos.com/calculator/sxg4x1lhf6

It's kind of crazy looking... but you can clearly see that the slope at the origin is -1.

Answer 55

weird, those y's are confusing tho

Answer 56

still need to figure out derivatives "with respect to".

its the "with respect to" part

Answer 57

im wondering about e^y now

Answer 58

\[\Large\rm \left(e^x\right)'=e^x~x'=e^x(1)\]

Answer 59

\[\Large\rm (e^y)'=e^y~y'\]And you would stop right there :)
yah?

Answer 60

ya?

Answer 61

just tried doing this one on my own
cos(y^2)+x=e^y

i got y'=1/(e^y+2ysin(y^2))

Answer 62

Mmm yes, looks good! :)