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help please: Use implicit differentiation to find the equation of the tangent line to the curve at the point. (attached below)

Mathematics
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i know i have to use implicit differentiation, then plug into derivative to find slope, plug slope and given points to point-slope formula.
Problem I have is step 1

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Other answers:

  • phi
take the derivative with respect to x. use the product rule for example, on the xy term you do this \[ \frac{d}{dx}xy = x \frac{d}{dx}y+ y \frac{d}{dx}x \\ = x \frac{dy}{dx}+ y \frac{dx}{dx} \\ = x \frac{dy}{dx}+ y\] use the same idea on \(xy^3\)
i got x(3y*dy/dx) +y^3
so.. 3xy*dydx +y^3
dy/dx**
  • phi
you mean 3 x y^2 dy/dx for the first part d/dx of y^3 is 3 y^2 dy/dx
okay im lost with the second part.
i know that you took the derivative of xy^3= 3xy^2
+xy= 1?
*dy/dx
  • phi
from the beginning \[ \frac{d}{dx}\left(xy^3 + xy = 4\right) \\ x \frac{d}{dx}y^3 + y^3 \frac{d}{dx}x+ x\frac{d}{dx}y+ y \frac{d}{dx}x=\frac{d}{dx}4\] that simplifies to \[ x \frac{d}{dx}y^3 + y^3+ x\frac{dy}{dx}+ y =0\] now you can do the d/dx of y^3 , right ?
yes.
  • phi
use the power rule \[ \frac{d y^n}{dx} = n y^{n-1} \frac{dy}{dx} \]
this isn't applying chain rule or is it? because i understand what youre doing here makes sense/
  • phi
yes, it is the chain rule
  • phi
normally you do d/dx of x^3 and you get 3x^2 dx/dx but the dx/dx is left off and the rule looks like \[ \frac{d\ x^3}{dx} = 3 x^2 \]
okay, let me give this a try.
okay I got this down: x*3y^2(dy/dx) +3y +x(dy/dx)+y=0
  • phi
you mean y^3 not 3y , right ?
oh so only find the derivative of [y^3]...
thanks btw! :)
then i should just plot the points for x and y
wait a min. what should i do with dy/dx?
  • phi
solve for dy/dx and then sub in for x and y to get the value of dy/dx at the point (2,1)
how?
  • phi
use algebra
question: is it possible to divide this: -4x/-2y ????
how the hell can someone divide this and get 2x/y?????
this is so frustrating, please help me understand this.
  • phi
what part is confusing ? you have \[3 x y^2\frac{dy}{dx} + y^3+ x\frac{dy}{dx}+ y =0 \] plug in x=2 and y =1 and solve for dy/dx (which is the slope)
  • phi
at the point (2,1)
no no, i know that. i was just looking at an example and i see that they divide -4x/-2y and got 2x/y
it frustrated me to see this and not know how they even possibly do it.
  • phi
are you asking how to get from \[ \frac{-4x}{-2y} = \frac{2x}{y} \] ? -4/-2 is +2
yes but with diferet variables?? 0.o
different*
  • phi
what is the question ?
oh the slope is -1/4! =]
  • phi
yes. also, the tangent line goes through point (2,1) so you can find the intercept.
cool, how do you know this?
  • phi
see http://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/implicit-differentiation-1 and the following videos , especially http://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/finding-slope-of-tangent-line-with-implicit-differentiation
okie doke
thanks @phi you helped me so much! ^_^ so thankful
  • phi
they asked you to find the tangent line through point (2,1) you found its slope = -1/4 you know it goes through 2,1 so in point-slope form you have y- 1 = (-1/4)(x-2) which you can change to slope-intercept form
y=-1/4x+3/2
  • phi
yes

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