I'm trying to force Mathematica to implicitly differentiate an ellipse equation of the form:

```
x^2/a^2+y^2/b^2 == 100
```

with `a = 8`

and `b = 6`

.

The command I'm using looks like this:

```
D[x^2/a^2 + y^2/b^2 == 100/. y -> 3/4*Sqrt[6400-x^2], x]
```

where, `y->3/4*Sqrt[6400-x^2]`

comes from solving `y`

in terms of `x`

.

I got this far by following the advice found here: http://www.hostsrv.com/webmaa/app1/MSP/webm1010/implicit

Input for this script is the conventional way that an implicit relationship beween x and y is expressed in calculus textbooks. In Mathematica you need to make this relationship explicit by using y[x] in place of y. This is done automatically in the script by replacing all occurances of y with y[x].

But the solution Mathematica gives does not have `y'`

or `dy/dx`

in it (like when I solved it by hand). So I don't think it's been solved correctly. Any idea on what command would get the program to solve an implicit differential? Thanks.

The conceptually easiest option (as you mentioned) is to make `y`

a function of `x`

and use the partial derivative operator `D[]`

```
In[1]:= D[x^2/a^2 + y[x]^2/b^2 == 100, x]
Solve[%, y'[x]]
Out[1]= (2 x)/a^2 + (2 y[x] y'[x])/b^2 == 0
Out[2]= {{y'[x] -> -((b^2 x)/(a^2 y[x]))}}
```

But for more complicated relations, it's best to use the total derivative operator `Dt[]`

```
In[3]:= SetOptions[Dt, Constants -> {a, b}];
In[4]:= Dt[x^2/a^2 + y^2/b^2 == 100, x]
Solve[%, Dt[y, x]]
Out[4]= (2 x)/a^2 + (2 y Dt[y, x, Constants -> {a, b}])/b^2 == 0
Out[5]= {{Dt[y, x, Constants -> {a, b}] -> -((b^2 x)/(a^2 y))}}
```

Note that it might be neater to use `SetAttributes[{a, b}, Constant]`

instead of the `SetOptions[Dt, Constants -> {a, b}]`

command... Then the `Dt`

doesn't carry around all that extra junk.

The final option (that you also mentioned) is to solve the original equation for `y[x]`

, although this is not always possible...

```
In[6]:= rep = Solve[x^2/a^2 + y^2/b^2 == 100, y]
Out[6]= {{y -> -((b Sqrt[100 a^2 - x^2])/a)}, {y -> (b Sqrt[100 a^2 - x^2])/a}}
```

And you can check that it satisfies the differential equation we derived above for both solutions

```
In[7]:= D[y /. rep[[1]], x] == -((b^2 x)/(a^2 y)) /. rep[[1]]
Out[7]= True
```

You can substitute your values `a = 8`

and `b = 6`

anytime with replacement rule `{a->8, b->6}`

.

If you actually solve your differential equation `y'[x] == -((b^2 x)/(a^2 y[x])`

using DSolve with the correct initial condition (derived from the original ellipse equation) then you'll recover the solution for `y`

in terms of `x`

given above.

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