Question

Use implicit differentiation to find an equation of the tangent line to the graph at the given point.
x + y − 1 = ln(x^10 + y^17), (1, 0)

Answer 1

if it's like the other question then it would be y = -x +1 +ln(x^10 + y^17)?

Answer 2

then derivation would be d/dx(-x) d/dx(1), ln(x^10 + y^17) and d/dx(x^10 + y^17) because of the chain rule?

Answer 3

which should be dy/dx = -1 + \[\frac{10x^9}{x^{10}} + \frac{17x^{16}}{x^{17}}\] +\[10x^9 + 17y^{16}\] ?

Answer 4

let's just look at the left side first
what is the derivative with respect to x of that?

Answer 5

I left y alone on the right, should I have no done that?

Answer 6

if I didn't it would be 1+ dy/dx

Answer 7

that's right :)
so we have 1+y'= ???
give the derivative of the RHS a try

Answer 8

I posted what I thought up above, is that not right?

Answer 9

well you have it all listed out so it's hard to tell
looks like you turned a y into x somehow though....

Answer 10

yeah I did oops :P

Answer 11

\[D_x(\ln u)=\frac{u'}u\]in your case\[u=x^{10}+y^{17}\]so what is \(u'\) ?

Answer 12

\[10x^9 + 17y^{16} ?\]

Answer 13

almost, but you forgot the most important part of implicit differentiation:
because y is a function of x, we have to use the chain rule on y, so we get\[10x^9+17y^{16}y'\]that y' (same thing as dy/dx, different notation) is what we are after. Omitting it will give the wrong answer.

Answer 14

so it's only for y, and only used for implicit differentiation? What exactly does that term mean?

Answer 15

and it just stays as y'? no function?

Answer 16

y is f(x) in disguise
so we could write this y(x)
when you take the derivative of some function of y(x) with respect to x that means you have to use the chain rule
for example if y(x) is raised to some power the chain rule with respect to x gives\[\frac d{dx}[y(x)]^n=n[y(x)]^{n-1}\cdot y'(x)\]

Answer 17

\[y'=\frac{dy}{dx}\]they are the same thing, it's just shorthand

Answer 18

ic... interesting. So if we have something with respect to something else, no matter what it is we have to use the chain rule.

Answer 19

it coudl be y, x, t, u, etc?

Answer 20

yes, we can always work our way inward toward the dependent variable we are taking the derivative with respect to
really this is\[u=y^n\]\[y=\ln t\]then\[\frac{du}{dt}=\frac{du}{dy}\cdot\frac{dy}{dt}=ny^{n-1}\cdot\frac1t=n(\ln t)^{n-1}\cdot\frac1t\]you can always sub back in for y (or whatever intermediate variable you have)at the end

Answer 21

in the case of your problem, however, subbing in for y in terms of x would be pointless and painful, so we won't do it this time

Answer 22

okies :)

Answer 23

so now we have\[1+y'={10x^9+17y^{16}y'\over x^{10}+y^{17}}\]do you agree?

Answer 24

yuh.. At first I guiess I got confused when I split x and y apart, I guess they always stay together as 'u' within ln() and other situations... :p

Answer 25

exactly, they stay together because they are the "u" in our formula

Answer 26

so the fancy way to do this is to solve it for y' (which is the slope remember), then plug in the point (1,0) and see what we get
however we could also be lazier and just plug in the numbers right now ;)
that should make solving for y' a lot easier

Answer 27

so it would be 10(1)^9/1^10

Answer 28

so 10/1 = 10 = 1 + y' ....... y' = 10-1 = 9

Answer 29

yep :)

Answer 30

didnt' do the other part since it's all 0.

Answer 31

now we can apply point slope form (hope you remember it) from algebra\[y-y_0=m(x-x_0)\]

Answer 32

so now we do the y-y0 = m(x-xo)?

Answer 33

echoechocecho....

Answer 34

ok so that's y-0 = 9(x-1) = y = 9x-9

Answer 35

and we're done :D

Answer 36

:D

Answer 37

so when would we solve for y'?

Answer 38

also if lets say y was 1 and x was 0 we would have 17(1)^16/(1)^17 * y' right?

Answer 39

we used y' already, it was our m, remember?

Answer 40

oh okay yeah.

Answer 41

right to your last post

Answer 42

you were saying before that was the "fancy way."

Answer 43

do the y's just cancel out?

Answer 44

no, if we \(don't\) plug in the numbers right away, and try to solve for y', that was what I meant by the fancy way, because it is harder and gives a more general result

the y' will not cancel out, it should be factorable if you do it right

Answer 45

ok because it's 1+y' = (all that stuff) *y' was just curious what happens since there are 2 of them.

Answer 46

if we plug in for x and y that doesn't affect the y' on the right , right?

Answer 47

you would get\[1+y'={10x^9+17y^{16}y'\over x^{10}+y^{17}}\]\[x^{10}+y^{17}+(x^{10}+y^{17})y'=10x^9+17y^{16}y'\]\[x^{10}+y^{17}-10x^9=17y^{16}y'-(x^{10}+y^{17})y'\]

Answer 48

but maybe y is always 0 so that explains why the y' cancels out on the right because you're multiplying by 0 but iDk :p

Answer 49

\[x^{10}+y^{17}-10x^9=[17y^{16}-(x^{10}+y^{17})]y'\]\[y'={x^{10}+y^{17}-10x^9\over17y^{16}-x^{10}-y^{17}}\]notice this gives 9 also, so we're good :)

Answer 50

the factoring out of the y' can be a little tricky sometimes in these things, so try to focus on that part

Answer 51

y is not always 0, we cannot treat y as zero until we are ready to solve for y'

Answer 52

hmp... so if we just go in and substitute we get rid of the y' on the right? I'm just confused where it goes... I see in this case it's gone because we multiply it by 0, but what if we don't?

Answer 53

then we could solve it by simple arithmetic
say we had
3+y'=y^3y'+x
at point (1,2)
we could just plug in the numbers now and get
3+y'=8y'+1
7y'=2
y'=2/7
or something like that

Answer 54

if, somehow, y' did cancel out (like if we had the above evaluated at (1,1) that would tell us that the derivative is undefined there I think... actually I'm not sure, but it doesn't happen often in problems I encounter

Answer 55

\[y^{3y'}\] ????

Answer 56

sorry for my lack of clarity\[3+y'=y^3y'+x\]at (1,2) becomes\[3+y'=(2^3)y'+1\]\[3+y'=8y'+1\]\[7y'=2\]\[y'=2/7\]

Answer 57

ah i got it haha. I was konfusing myself with this calculus stuff I forgot about basic math.. I kept thinking it would just be y' all the time iand it would just cancel itself out or be cancelled by 0, thanks :).

Answer 58

welcome, good luck :)

Answer 59

:D on to the next question!