Thread: Implicit differentiation - not defined?

I have the problem:

Find the linear approximation about (-1,1) to the implicit function given by the relation: 2xy-x⁸+y²=-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)?

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.

Hm, I always used the formula:

x-x₀= x'/y' (x-x₀)

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?

the tangent line is x = -1

Originally Posted by infernalmich

Hm, I always used the formula:

x-x₀= x'/y' (x-x₀)

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

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.

The function in my example is given implicitly:
2xy-x⁸+y²=-2
find the tangent line at the point: (-1,1) - I refer to this points as x₀ and y₀

hence y'_(x)=- (af/ax) / (af/ay)
so y=y₀-y'_(x) (x-x₀)

In this case y=1-0
so y=1

____

Trying to do the same with:
x=x₀-x'_(y)(y-y₀)
x=-1- 0*(y-1)

now my stupid question:
Why is x=-1 the tangent line and not y=1?

Originally Posted by infernalmich

The function in my example is given implicitly:
2xy-x⁸+y²=-2
find the tangent line at the point: (-1,1) - I refer to this points as x₀ and y₀

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.

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:
2x³y³-3x²y³=5

af/ax=-5
af/ay=10

than I can calculate:
y=y⁰-(-2)(x-x⁰)
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?

Originally Posted by infernalmich

Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:
2x³y³-3x²y³=5

the point is not on the curve

did you copy the problem correctly?

no I didn't, sorry.

The correct problem is:
2x³y²-3x²y³=5

Originally Posted by infernalmich

no I didn't, sorry.

The correct problem is:
2x³y²-3x²y³=5

I get ...

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

Thank you!

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

find the tangent line to the equation:

y²+e^2x+xln(y)=4 through (0,2)

First I take the derivatives:

f_y(x,y)=2e²+y²+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 :-)

Originally Posted by infernalmich

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

find the tangent line to the equation:

y²+e^2x+xln(y)=4 through (0,2)

First I take the derivatives:

f_y(x,y)=2e²+y²+ln(y)

This is wrong to start with. f_y= 2y+ x/y.

f_x(x,y)=2y*e^2x+x/y

No, f_x= 2e^2x+ ln(y).

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.

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!

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

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

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.

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?