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

- anonymous

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- anonymous

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- anonymous

i know i have to use implicit differentiation, then plug into derivative to find slope, plug slope and given points to point-slope formula.

- anonymous

Problem I have is step 1

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## More 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 \(xy^3\)

- anonymous

i got x(3y*dy/dx) +y^3

- anonymous

@phi

- anonymous

so.. 3xy*dydx +y^3

- anonymous

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

- anonymous

okay im lost with the second part.

- anonymous

i know that you took the derivative of xy^3= 3xy^2

- anonymous

+xy= 1?

- anonymous

*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 ?

- anonymous

yes.

- phi

use the power rule
\[ \frac{d y^n}{dx} = n y^{n-1} \frac{dy}{dx} \]

- anonymous

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
\[ \frac{d\ x^3}{dx} = 3 x^2 \]

- anonymous

okay, let me give this a try.

- anonymous

okay I got this down: x*3y^2(dy/dx) +3y +x(dy/dx)+y=0

- phi

you mean y^3 not 3y , right ?

- anonymous

oh so only find the derivative of [y^3]...

- anonymous

thanks btw! :)

- anonymous

then i should just plot the points for x and y

- anonymous

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)

- anonymous

how?

- phi

use algebra

- anonymous

question: is it possible to divide this: -4x/-2y ????

- anonymous

how the hell can someone divide this and get 2x/y?????

- anonymous

this is so frustrating, please help me understand this.

- anonymous

@asnaseer

- 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)

- anonymous

no no, i know that. i was just looking at an example and i see that they divide -4x/-2y and got 2x/y

- anonymous

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

- anonymous

yes but with diferet variables?? 0.o

- anonymous

different*

- phi

what is the question ?

- anonymous

oh the slope is -1/4! =]

- phi

yes. also, the tangent line goes through point (2,1)
so you can find the intercept.

- anonymous

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

- anonymous

okie doke

- anonymous

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

- anonymous

y=-1/4x+3/2

- phi

yes

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