tangent to x^3-y^3+4xy = 1

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dangerous_dave

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First of all, this is another question, but is it right?

let y=(x^3_1)^(x^2). Find y'(1)

y'=(x^2(x^3+1)^(x^2 - 1))(3x^2)
y'(1)=(1(2)^0)(3)

y'(1) = 3

Thats my answer, and I think (hope) its right.

Anyway, on the the real question:

Find the equation of the tangent to the curve defined by the equation:

x^3 - y^3 + 4xy = 1

at the point (1,2)

Now I know I have to differentiate it, and I think that means I have to rearrange it to get y = something. Thing is, I can't seem to get it y = something without another y.
Best I've managed is :

y = cubed root of (x^3 + 4xy +1)

Please help :)
 
dangerous_dave said:
First of all, this is another question, but is it right?

let y=(x^3_1)^(x^2). Find y'(1)<<<what is that? from your work - it probably should be '+'

y'=(x^2(x^3+1)^(x^2 - 1))(3x^2)<<< No...No..Double no

y = (x^3 + 1)^(x^2)

ln(y) = x^2 * ln(x^3 + 1)

1/y * dy/dx = 2x*ln(x^3 + 1) + x^2 * 1/(x^3 + 1) * 3x^2

Now clean it up....

You cannot use that rule [d/dx(x^n) = n*x^(n-1)] when n is variable like 'x^2'
y'(1)=(1(2)^0)(3)

y'(1) = 3

Thats my answer, and I think (hope) its right.

Anyway, on the the real question:

Find the equation of the tangent to the curve defined by the equation:

x^3 - y^3 + 4xy = 1

at the point (1,2)

Now I know I have to differentiate it, and I think that means I have to rearrange it to get y = something. Thing is, I can't seem to get it y = something without another y.
Best I've managed is :

y = cubed root of (x^3 + 4xy +1)

You would need to differentiate implicitly - look up the procedure in your text book or do a google search

If you are still stuck write back showing us what you have found.

Please help :)
Click to expand...
 
Thanks for the help :)

For the second, I now have this:

y' = (-4y - 3x^2)/(-3y^2+4x)

Thats differentiated (I think). And now I just need to get y'(1), which I can do if the first bit is right :D
 
X^3-y^3+4xy=1 y' = (-4y-3x^2)/(-3y^2+4x) Point = (1,2) when x=1 and y=2, then y' = m = 11/8

We have point (1,2) and m =11/8. Ergo y-2 = (11/8)(x-1), y = (11x+5)/8.

I think your problem is that your y' is incorrect.
 
dangerous_dave said:
Could you please explain how you got that?
Click to expand...
Would you kindly please show your work? (It is much easier to help students find their mistakes when we can see those mistakes.)

Thank you for your consideration.

Eliz.
 
Ok I got this:

3x^2 - 3y^2y' + 4(xy' + (1)y) = 0

3x^2 - 3y^2y' + 4xy' + 4y = 0

-3y^2y' + 4xy' = -4y - 3x^2

y' (-3y^2 + 4x) = -4y - 3x^2

y' = (-4y - 3x^2)/(-3y^2 + 4x)


I just followed an example one that I found on the net, using my numbers instead of theirs. So where did I go wrong?


Wait, I jst realised we got the same y'! Ok so I get that y' = 11/8, but whats with y = (11x+5)/8?

Edit: Sorry I understand now, the equation is at 1,2! Thanks for your help :D
 
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