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.

- anonymous

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- myininaya

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

- anonymous

3x^2

- myininaya

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

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## More answers

- anonymous

3y^2

- myininaya

Now exactly

- myininaya

Not*

- anonymous

3y^2 y'

- myininaya

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

- myininaya

Solve for y'

- anonymous

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

- myininaya

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

- myininaya

See how I got my y' terms together

- anonymous

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

- myininaya

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

- myininaya

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

- myininaya

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

- anonymous

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

- myininaya

yes 3 for x and -2 for y

- anonymous

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

- myininaya

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

- anonymous

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=

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

You try (xy)'

- anonymous

1*y+xy'

- myininaya

ok great!

- myininaya

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

- anonymous

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

- myininaya

yep :)

- anonymous

m=-1/14
woooo thanks again :)

- myininaya

all you! :)

- anonymous

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.

- anonymous

3^x+log_2(xy)=10

- myininaya

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

- anonymous

where did the ln go in
(ln(xy))′

- myininaya

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

- anonymous

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

- myininaya

(xy)'=y+xy'

- anonymous

Ahh oops, I just saw that

- myininaya

Everything else looks good :)

- myininaya

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

- anonymous

why did you take the ln of 3^x?

- myininaya

\[a>0,(a^x)'=\ln(a) \cdot a^x\]
\[y=a^x\]
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
\[y'=a^x \cdot \ln(a) \]

- anonymous

Okay, I see it now
ln(3)⋅3x+1ln(2)⋅y+xy′xy=0

- anonymous

3^x*

- myininaya

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

- anonymous

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

- myininaya

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'

- myininaya

that include the factor y'*

- myininaya

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

- myininaya

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

- anonymous

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

- myininaya

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

- anonymous

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

- myininaya

Perfect! :)

- anonymous

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

- myininaya

Yep yep Great job kolo!

- myininaya

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.

- anonymous

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

- myininaya

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

- anonymous

Haha indeed it was, thanks again :)

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