Question

Use implicit differentiation to find the slope of the tangent line to the curve
x^3+y^3=y+21
at the point ( 3 , -2 ) and use it to find the equation of the tangent line in the form y=mx+b.

Answer 1

\[(x^3)'=?\]

Answer 2

3x^2

Answer 3

\[(y^3)'=?\]

Answer 4

3y^2

Answer 5

Now exactly

Answer 6

Not*

Answer 7

3y^2 y'

Answer 8

\[3x^2+3y^2y'=y'+0\]

Answer 9

Solve for y'

Answer 10

(1-3x^2)/(3y^2)

Answer 11

\[3yy'-y'=-3x^2\]

Answer 12

See how I got my y' terms together

Answer 13

yes, and then y' turns into -1 or you do put it in front like y'[3y]

Answer 14

I did that so I can factor out y' and divide by whatever it is being multiplied by

Answer 15

\[y'(3y^2-1)=-3x^2\]

Answer 16

\[y'=\frac{-3x^2}{3y^2-1}\]

Answer 17

and then i just plug in (3,-2) for x and y?

Answer 18

yes 3 for x and -2 for y

Answer 19

Thank you very much, the answer was y=-27/11x+59/11

Answer 20

Great stuff! :)
I'm glad you got it!

Answer 21

another one that was giving me trouble was this one

Use implicit differentiation to find the slope of the tangent line to the curve
xy^3+xy= 14
at the point ( 7 , 1 ).
m=

Answer 22

\[(xy^3)'=? \]
You need product rule and chain rule

Answer 23

\[=(x)'y^3+x(y^3)'\]

Answer 24

\[(xy^3)'=1 \cdot y^3+x \cdot 3y^2 y'\]

Answer 25

You try (xy)'

Answer 26

1*y+xy'

Answer 27

ok great!

Answer 28

so we have
\[ y^3+x \cdot 3y^2 y'+y+xy'=0\]

Answer 29

is it
y'=(-y^3-y)/(3xy^2+x)

Answer 30

yep :)

Answer 31

m=-1/14
woooo thanks again :)

Answer 32

all you! :)

Answer 33

Here's another where it uses log and I don't know what to do.

Use implicit differentiation to find the slope of the tangent line to the curve
3^x+\log_2(xy)=10
at the point ( 2 , 1 ) and use it to find the equation of the tangent line in the form y=mx+b.

Answer 34

3^x+log_2(xy)=10

Answer 35

\[(a^x)'=\ln(a) \cdot a^x\]
a>0
----
\[(log_2(xy))'=(\frac{\ln(xy)}{\ln(2)})'=\frac{1}{\ln(2)} \cdot (\ln(xy))'=\frac{1}{\ln(2)} \cdot \frac{(xy)'}{xy}\]

Answer 36

where did the ln go in

(ln(xy))′

Answer 37

(ln(f))'=f'/f
(ln(inside))'=derivative of inside/inside

Answer 38

Is this right?
1/ln(2) * (1*y'+x*1)/(xy)

Answer 39

(xy)'=y+xy'

Answer 40

Ahh oops, I just saw that

Answer 41

Everything else looks good :)

Answer 42

So we have:
\[\ln(3) \cdot 3^x+\frac{1}{\ln(2)} \cdot \frac{y+xy'}{xy}=0\]
agree?

Answer 43

why did you take the ln of 3^x?

Answer 44

\[a>0,(a^x)'=\ln(a) \cdot a^x\]
\[y=a^x\]
Take ln( ) of both sides!
\[\ln(y)=\ln(a^x)\]
Use one of your properties to rewrite
\[\ln(y)=x \ln(a)\]
Now differentiate both sides!
\[\frac{y'}{y}=\ln(a)\]
Multiply y on both sides to isolate y'
So we have
\[y'= y \ln(a)\]
\[ \text{ recall from the beginning that } y=a^x\]
So we have
\[y'=a^x \cdot \ln(a) \]

Answer 45

Okay, I see it now

ln(3)⋅3x+1ln(2)⋅y+xy′xy=0

Answer 46

3^x*

Answer 47

\[\ln(3) \cdot 3^x+\frac{1}{\ln(2)} \cdot \frac{y+xy'}{xy}=0 ?\]

Answer 48

yes that, apparently copy paste doesn't work..it left out the division signs

Answer 49

Ok so we need to isolate y' , right?
I would start by putting terms without the factor y' on the opposite side of terms the include the factor y'

Answer 50

that include the factor y'*

Answer 51

Like so...
\[\frac{1}{\ln(2)} \cdot \frac{y+xy'}{xy}=-\ln(3) \cdot 3^x\]

Answer 52

Now multiply both sides by the left hand's bottom's junk!

Answer 53

like this?
y+xy'=(-ln(3)*3^x)/(ln(2)xy)

Answer 54

You multiplied on one side but divided on the other by the junk I told you to multiply on both sides

Answer 55

y+xy'=(-ln(3)*3^x)(ln(2)xy)

Answer 56

Perfect! :)

Answer 57

y'=((-ln(3)*3^x)(ln(2)xy)-y)/(x)

Answer 58

Yep yep Great job kolo!

Answer 59

Ok hey I have to go
Do you think you got it from here?
If not, please post a link in the chats? I'm sorry.

Answer 60

Yes, I think I can take it from here, thanks so much for all the help!

Answer 61

No problem. Sorry I have to go :(
It was fun differentiating stuff with you!

Answer 62

Haha indeed it was, thanks again :)