anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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myininaya
  • myininaya
\[(x^3)'=?\]
anonymous
  • anonymous
3x^2
myininaya
  • myininaya
\[(y^3)'=?\]

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anonymous
  • anonymous
3y^2
myininaya
  • myininaya
Now exactly
myininaya
  • myininaya
Not*
anonymous
  • anonymous
3y^2 y'
myininaya
  • myininaya
\[3x^2+3y^2y'=y'+0\]
myininaya
  • myininaya
Solve for y'
anonymous
  • anonymous
(1-3x^2)/(3y^2)
myininaya
  • myininaya
\[3yy'-y'=-3x^2\]
myininaya
  • myininaya
See how I got my y' terms together
anonymous
  • anonymous
yes, and then y' turns into -1 or you do put it in front like y'[3y]
myininaya
  • myininaya
I did that so I can factor out y' and divide by whatever it is being multiplied by
myininaya
  • myininaya
\[y'(3y^2-1)=-3x^2\]
myininaya
  • myininaya
\[y'=\frac{-3x^2}{3y^2-1}\]
anonymous
  • anonymous
and then i just plug in (3,-2) for x and y?
myininaya
  • myininaya
yes 3 for x and -2 for y
anonymous
  • anonymous
Thank you very much, the answer was y=-27/11x+59/11
myininaya
  • myininaya
Great stuff! :) I'm glad you got it!
anonymous
  • 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
  • myininaya
\[(xy^3)'=? \] You need product rule and chain rule
myininaya
  • myininaya
\[=(x)'y^3+x(y^3)'\]
myininaya
  • myininaya
\[(xy^3)'=1 \cdot y^3+x \cdot 3y^2 y'\]
myininaya
  • myininaya
You try (xy)'
anonymous
  • anonymous
1*y+xy'
myininaya
  • myininaya
ok great!
myininaya
  • myininaya
so we have \[ y^3+x \cdot 3y^2 y'+y+xy'=0\]
anonymous
  • anonymous
is it y'=(-y^3-y)/(3xy^2+x)
myininaya
  • myininaya
yep :)
anonymous
  • anonymous
m=-1/14 woooo thanks again :)
myininaya
  • myininaya
all you! :)
anonymous
  • 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
  • anonymous
3^x+log_2(xy)=10
myininaya
  • 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
  • anonymous
where did the ln go in (ln(xy))′
myininaya
  • myininaya
(ln(f))'=f'/f (ln(inside))'=derivative of inside/inside
anonymous
  • anonymous
Is this right? 1/ln(2) * (1*y'+x*1)/(xy)
myininaya
  • myininaya
(xy)'=y+xy'
anonymous
  • anonymous
Ahh oops, I just saw that
myininaya
  • myininaya
Everything else looks good :)
myininaya
  • myininaya
So we have: \[\ln(3) \cdot 3^x+\frac{1}{\ln(2)} \cdot \frac{y+xy'}{xy}=0\] agree?
anonymous
  • anonymous
why did you take the ln of 3^x?
myininaya
  • 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
  • anonymous
Okay, I see it now ln(3)⋅3x+1ln(2)⋅y+xy′xy=0
anonymous
  • anonymous
3^x*
myininaya
  • myininaya
\[\ln(3) \cdot 3^x+\frac{1}{\ln(2)} \cdot \frac{y+xy'}{xy}=0 ?\]
anonymous
  • anonymous
yes that, apparently copy paste doesn't work..it left out the division signs
myininaya
  • 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
  • myininaya
that include the factor y'*
myininaya
  • myininaya
Like so... \[\frac{1}{\ln(2)} \cdot \frac{y+xy'}{xy}=-\ln(3) \cdot 3^x\]
myininaya
  • myininaya
Now multiply both sides by the left hand's bottom's junk!
anonymous
  • anonymous
like this? y+xy'=(-ln(3)*3^x)/(ln(2)xy)
myininaya
  • myininaya
You multiplied on one side but divided on the other by the junk I told you to multiply on both sides
anonymous
  • anonymous
y+xy'=(-ln(3)*3^x)(ln(2)xy)
myininaya
  • myininaya
Perfect! :)
anonymous
  • anonymous
y'=((-ln(3)*3^x)(ln(2)xy)-y)/(x)
myininaya
  • myininaya
Yep yep Great job kolo!
myininaya
  • 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
  • anonymous
Yes, I think I can take it from here, thanks so much for all the help!
myininaya
  • myininaya
No problem. Sorry I have to go :( It was fun differentiating stuff with you!
anonymous
  • anonymous
Haha indeed it was, thanks again :)

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