Implicit differentiation - not defined?

I have the problem:

Find the linear approximation about (-1,1) to the implicit function given by the relation: 2xy-x^{8}+y^{2}=-2

My problem is that trying implicit differentiation, I always get a slope which is not defined, because the denumerator is always (x+y) and therefore 0.

Did I do an error doing my calculation? If my calculation was right, can I answer that the tangent line is not defined at (-1,1)?

Re: Implicit differentiation - not defined?

No, there is no error. You can, however, say that the "linear approximation" is the vertical line x= -1.

Another way of looking at it is to treat this as an implicitly defining x as a function of y. Then, differentiating both sides with respect to y you would get x'= 0. That gives x= 0(y-1)- 1 or, again, x= -1.

Re: Implicit differentiation - not defined?

Hm, I always used the formula:

x-x_{0}= x'/y' (x-x_{0})

In that case for this example I get:

y-1=0*(x+1)

so

y=1

what was my mistake? Did I use the wrong formula in order to get the tangent line?

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Re: Implicit differentiation - not defined?

the tangent line is x = -1

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by

**infernalmich** Hm, I always used the formula:

x-x_{0}= x'/y' (x-x_{0})

What do x' and y' **mean** here? (And surely, you don't mean on both sides?)

Quote:

In that case for this example I get:

y-1=0*(x+1)

so

y=1

what was my mistake? Did I use the wrong formula in order to get the tangent line?

I can't answer that until I know what you mean by x' and y'. Normally, I would interpret them as derivatives, but with respect to what variables? If x and y were given as functions of some other variable, and x' and y' were the derivatives with respect to that third variable, the tangent line would be given by where the slope is the reciprocal of what you give.

Re: Implicit differentiation - not defined?

The function in my example is given implicitly:

2xy-x^{8}+y^{2}=-2

find the tangent line at the point: (-1,1) - I refer to this points as x_{0} and y_{0}

hence y'_{(x)}=- (af/ax) / (af/ay)

so y=y_{0}-y'_{(x)} (x-x_{0})

In this case y=1-0

so y=1

____

Trying to do the same with:

x=x_{0}-x'_{(y)}(y-y_{0})

x=-1- 0*(y-1)

now my stupid question:

Why is x=-1 the tangent line and not y=1?

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by

**infernalmich** The function in my example is given implicitly:

2xy-x^{8}+y^{2}=-2

find the tangent line at the point: (-1,1) - I refer to this points as x_{0} and y_{0}

now my stupid question:

Why is x=-1 the tangent line and not y=1?

first, is not a function ... why?

second, all points on a vertical line have the same x-value, hence, the equation of the vertical line passing thru the point (-1,1) is x = -1

fyi, all points on a horizontal line have the same y-value , hence the equation of a horizontal line would be y = a constant.

Re: Implicit differentiation - not defined?

Now I try with a second one:

Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:

2x^{3}y^{3}-3x^{2}y^{3}=5

af/ax=-5

af/ay=10

than I can calculate:

y=y^{0}-(-2)(x-x^{0})

y=-1+2*(x-1)

y=2x+2

0,5y=x

and for x:

x=1-(-0,5)(y--1))

x=0,5y

am I right?

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by

**infernalmich** Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:

2x^{3}y^{3}-3x^{2}y^{3}=5

the point is not on the curve

did you copy the problem correctly?

Re: Implicit differentiation - not defined?

no I didn't, sorry.

The correct problem is:

2x^{3}y^{2}-3x^{2}y^{3}=5

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Re: Implicit differentiation - not defined?

Quote:

Originally Posted by

**infernalmich** no I didn't, sorry.

The correct problem is:

2x^{3}y^{2}-3x^{2}y^{3}=5

I get ...

Re: Implicit differentiation - not defined?

Oh I understood my error. I messed up one of the - signs :-)

Thank you!

Re: Implicit differentiation - not defined?

I have another question. Could someone check this for me please:

find the tangent line to the equation:

y^{2}+e^{2x}+xln(y)=4 through (0,2)

First I take the derivatives:

f_{y}_{(x,y)}=2e^{2}+y^{2}+ln(y)

f_{x}[SUB](x,y)[/SUB=2y*e^{2x}+x/y

through the point (0,2) I have:

f_{y}=ln(2)+2

f_{x}=4

so my tangent line is:

y-2=-(ln(2)+2)/4*x

y=x/2-ln(2)+2

Is this true? I really hope so, because I am trying this examples since 2 hours :-)

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by

**infernalmich** I have another question. Could someone check this for me please:

find the tangent line to the equation:

y^{2}+e^{2x}+xln(y)=4 through (0,2)

First I take the derivatives:

f_{y}_{(x,y)}=2e^{2}+y^{2}+ln(y)

This is wrong to start with. f_{y}= 2y+ x/y.

Quote:

f_{x}(x,y)=2y*e^{2x}+x/y

No, f_{x}= 2e^{2x}+ ln(y).

Quote:

through the point (0,2) I have:

f_{y}=ln(2)+2

f_{x}=4

No f_{y}(0, 2)= 4 and f_{x}(0, 2)= 2+ ln(2). You have "x" and "y" reversed.

Quote:

so my tangent line is:

y-2=-(ln(2)+2)/4*x

Oddly enough, you have again mixed up "x" and "y" so that this **is** correct!

This, however, is not correct. You should have -((ln(2)+ 2)/4)x+ 2. The x is not just multiplied by the "2/4".

Quote:

Is this true? I really hope so, because I am trying this examples since 2 hours :-)

You really need to review basic differentiation rules and algebra.

Re: Implicit differentiation - not defined?

I have another question:

Find all values for which the tangent line of the following equation is horizontal:

first I get the derivatives with respect to x and y:

af/ax=4x+y

af/ay=x+2y

Than I use implicit differentiation: - (4x+y)/(x+2y)

As I remember, if the numerator is 0 the tangent line is horizontal, so the tangent line is horiziontal for y=4x or x=y/4 - if I would be asked for the vertical tangent line than the denumerator would have to be 0.

Is this true?