what is the slope of the tangent line to the curve (x^3 + y)^2 -2a^2xy = b^2 at the point P(0,1) when a=(2)^(1/2) and b = 1

- anonymous

- chestercat

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

- anonymous

I'm getting DNE but its wrong

- Jhannybean

For this problem can you input the values for a and b, then find the derivative? Or does the inputting of values happen after differentiating..

Looking for something else?

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

## More answers

- Jhannybean

It looks really ugly if you differentiate it along with a and b :\

- Jhannybean

I would use implicit differentiation to find the derivative.

- anonymous

i really have no idea , iv imputed the values for a and be and then solved for y and derived that , but that seems to be wrong

- Jhannybean

Yeah it is hahahahaha

- Jhannybean

Because that's basically stating the tangent line is found at a point before even being found.

- anonymous

you only need to find f'(x,y)

- anonymous

but before u start it says when a=(2)^(1/2) and b = 1 , a and b seems like small values and you can sub them why it mention them in ur question ? seems like there is a formula in ur book or something

- IrishBoy123

\[\vec n = <2(x^3+y)3x^2 - 2a^2y, 2(x^3+y)-2a^2x>\]
\[= <6x^2(x^3+y) - 2a^2y, 2(x^3+y)-2a^2x>\]
x = 0
\[= <-2a^2y,2>\]
\[y = 1, a = \sqrt{2}\]
\[\vec n = <-4,2,>\]

- freckles

don't solve for y
just find the derivative of both sides w.r.t. x
a and b are constants
you can input those values in later or not
once you have differentiated both sides
you can input the values for a and b if you haven't already
and then also input 0 for x and 1 for y
then solve for y' to find the slope.

- anonymous

see thats what i'm talking about :D like @irishboy123

- IrishBoy123

|dw:1441778557798:dw|

- anonymous

just noticed the name of the chapter is implicit differention

- freckles

why is there y'=that mess=0?

- freckles

wait
I just had a problem with y'= part

- anonymous

well the question said (tangent line) so it's line equation
u can do it in both ways with irish normal vector or with slope and a point

- freckles

\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }\]

- Jhannybean

Thanks.

- freckles

I just didn't agree with the y'=that stuff=0

- freckles

because saying y'=0 is saying you already know the slope to be 0

- freckles

\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \text{ was right }\]
so copying this
@Jdosio just enter in 0 for x and 1 for y
simplify and solve for y'

- anonymous

okay

- anonymous

a^2 ?

- freckles

yes and a was?

- anonymous

a = 2^(1/2) so final answer is 2

- freckles

right
but hey listen do you understand how she got that equation?

- anonymous

not a single clue ^_^

- freckles

she used chain rule and product rule
(there are some other small potato rules she used)

- freckles

\[\frac{d}{dx}(x^3+y)^2 \]
for this you should use chain rule

- Jhannybean

\[ (x^3 + y)^2 -2a^2xy = b^2\]\[2(x^3+y)(3x^2+y')-(2a^2xy'+2a^2y)=0 \]\[6x^5+2x^3y'+6x^3y+2yy'-2a^2xy'-2a^2y=0\]\[y'(2x^3+2y-2a^2x)=-6x^5 -6x^3y+2a^2y\]\[y'=...\]

- freckles

\[\frac{d}{dx}(x^3+y)^2 =2(x^3+y) \cdot (x^3+y)'\]
derivative of outside times the derivative of inside

- freckles

\[\frac{d}{dx}(x^3+y)=3x^2+y' \\ \text{ so } \frac{d}{dx}(x^3+y)^2=2(x^3+y) \cdot (3x^2+y')\]

- Jhannybean

chain rule: \(f(g(x)) = f'(x) \cdot g'(x)\)

- anonymous

oh okay i see i see

- anonymous

i was just rather lost on the second part of the subtraction where you used the product rule , im familiar with it i just wasn't entirely sure how to go about the a but i see what you did

- freckles

the other term you can use constant multiple rule and product rule
\[\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy)\]

- anonymous

yea
x' y + x y'

- Jhannybean

You're trying to keep in mind that taking the derivative of \(y\) will give you \(\dfrac{dy}{dx}\) and that you are taking the derivative of \(x\) as well.

- freckles

that is correct but x'=?

- freckles

or dx/dx=?

- anonymous

wait , are you asking about the example or the actual question?

- freckles

well you said
\[\frac{d}{dx}(xy)=\frac{dx}{dx} y+x \frac{dy}{dx}\]
and I was asking you to simplify dx/dx

- anonymous

well 1 right?

- freckles

yep

- freckles

\frac{d}{dx}-2a^2xy \\ -2a^2 \frac{d}{dx}(xy) \text{ by constant multiple rule } \\ \text{ do you think you can use product rule \to find } \frac{d}{dx}(xy)
right
\[\frac{d}{dx}(xy)=y+xy' \\ \text{ so } \frac{d}{dx}-2a^2xy=-2a^2(y+xy') =-2a^2y-2a^2xy'\]

- freckles

and of course the other side of the equation is b^2
so when differentiating a constant you get 0

- freckles

@Jdosio do you have any questions on this?

- anonymous

not on this question specifically , if i come up with anything else i missed ill ask you guys if your still around. thanks for all your help once again.

- Jhannybean

No problem.

Looking for something else?

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