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I need someone to explain to me step by step the following problem:
 one year ago
 one year ago
I need someone to explain to me step by step the following problem:
 one year ago
 one year ago

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bmelykBest ResponseYou've already chosen the best response.1
Find the equation of the tangent line to the curve \[x^{2}y^{2}1=3(x+y)\] at the points (1,1)
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
I have the problem worked out in front of me (prof gave us the answers) but, his work is disorganized and I don't understand what it is he's doing.
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
I figure I have to do this implicitly.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes  you are correct  use implicit differentiation to get an expression for slope of the tangent line at any point (x,y). then substitute x=1, y=1 to find the value of the slope of the tangent line at (1, 1) then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, 1)
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
that's not the answer.... the answer is: y=1/5x+4/5
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
okay and the way my prof did his differentiation was really weird. his first step looks like: x^{2}2y*y'+y^{2}(2x)=3(1+y')
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
that's not how i would start the implicit differentiation.
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
what your prof did is correct  he is using the product rule, see here: http://www.1728.org/chainrul.htm
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
how would you have done this?
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
haha no idea... i suck at math : (
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
so i use the product rule for x^2*y^2 ??
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
that is just straight differentiation since it can be written as: 3x + 3y
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
so that will just equal 3y' ??
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
so you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
okay so now i have the same first line as he does haha.
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
i ignored the 3 outside the brackets because it's a constant right?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes. thats good  you are making progress  maybe you are a lot better at maths than you think. :)
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
haha maybe, maybe not...
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
so now i can take the y^2(2x) over to the other side?
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
and the right hand side over to the left?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yes  what you need to do is to get an expression for y'. so move all terms involving y' to the left and the rest to the right
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
first expand the expression on the right
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
do i leave my 3 on the right hand side?
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
okay so now i have: (x^2)(2y*y')(1y')=3(2xy^2)
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
that look right to you?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
you started with this:\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]so first expand the right to get:\[x^{2}(2y*y')+y^{2}(2x)=3+3y'\]then move terms involving y' to the left and the rest to the right to get:\[x^{2}(2y*y')3y'=3y^{2}(2x)\]
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
then factorise y' on the left
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
\[y'=\frac{ 32xy^{2} }{ 2x^{2}y3 }\]
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
woooo. so now that part is done, right?
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
yup  the rest should be straight forward. I need to go for dinner now. I'll check back afterwards to see if you had any further problems.
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
the steps left are: then substitute x=1, y=1 to find the value of the slope of the tangent line at (1, 1) then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, 1)
 one year ago

bmelykBest ResponseYou've already chosen the best response.1
i got the rest of it now ; ) thanks a bunch!
 one year ago

asnaseerBest ResponseYou've already chosen the best response.1
thats great! and you are more than welcome! :)
 one year ago
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