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 $\displaystyle x- x_0$ 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 $\displaystyle y- y_0= (y'/x')(x- x_0)$ 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, $\displaystyle 2xy-x^8+y^2=-2$ 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 $\displaystyle (1,-1)$ is not on the curve $\displaystyle 2x^3y^3 - 3x^2y^3 = 5$

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 ...

$\displaystyle \color{red}{y + 1 = \frac{12}{13}(x - 1)}$

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:

$\displaystyle 2x^2+xy+y^2-4=-2$

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?