mathcalculus
help please: Use implicit differentiation to find the equation of the tangent line to the curve at the point. (attached below)
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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.
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Problem I have is step 1
phi
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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 \(xy^3\)
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i got x(3y*dy/dx) +y^3
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@phi
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so.. 3xy*dydx +y^3
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dy/dx**
phi
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you mean 3 x y^2 dy/dx for the first part
d/dx of y^3 is 3 y^2 dy/dx
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okay im lost with the second part.
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i know that you took the derivative of xy^3= 3xy^2
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+xy= 1?
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*dy/dx
phi
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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 ?
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yes.
phi
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use the power rule
\[ \frac{d y^n}{dx} = n y^{n-1} \frac{dy}{dx} \]
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this isn't applying chain rule or is it? because i understand what youre doing here makes sense/
phi
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yes, it is the chain rule
phi
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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
\[ \frac{d\ x^3}{dx} = 3 x^2 \]
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okay, let me give this a try.
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okay I got this down: x*3y^2(dy/dx) +3y +x(dy/dx)+y=0
phi
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you mean y^3 not 3y , right ?
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oh so only find the derivative of [y^3]...
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thanks btw! :)
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then i should just plot the points for x and y
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wait a min. what should i do with dy/dx?
phi
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solve for dy/dx and then sub in for x and y to get the value of dy/dx at the point (2,1)
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how?
phi
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use algebra
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question: is it possible to divide this: -4x/-2y ????
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how the hell can someone divide this and get 2x/y?????
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this is so frustrating, please help me understand this.
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@asnaseer
phi
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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
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at the point (2,1)
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no no, i know that. i was just looking at an example and i see that they divide -4x/-2y and got 2x/y
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it frustrated me to see this and not know how they even possibly do it.
phi
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are you asking how to get from
\[ \frac{-4x}{-2y} = \frac{2x}{y} \]
?
-4/-2 is +2
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yes but with diferet variables?? 0.o
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different*
phi
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what is the question ?
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oh the slope is -1/4! =]
phi
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yes. also, the tangent line goes through point (2,1)
so you can find the intercept.
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cool, how do you know this?
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okie doke
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thanks @phi you helped me so much! ^_^ so thankful
phi
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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
mathcalculus
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y=-1/4x+3/2
phi
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yes