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

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- anonymous

@ganeshie8 @Robert136 @nincompoop

- 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..

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## 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.

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