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

- anonymous

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

##### 1 Attachment

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.