# Math Help - Differentiating Implicit Equations

1. ## Differentiating Implicit Equations

If
Code:
sin^4(\[Alpha])/x+cos^4(\[Alpha])/y-1/(x+y)==0
^^^^(it looked fine in mathematica...can someone tell me how to write code in this forum?)

then what is
Code:
dy/dy??
*here [alpha] is a constant*

well i figured out dy/dx and it came out to be
Code:
y/x
but the book says
Code:
dy/dx=tan^2[Alpha]
,which is a constant. Does this mean the above line represents a straight line??Mathematica plotted a pair of lines(i guess) and i cannot ask mathematica to find
Code:
 dy/dx
(if its possible to find (dy/dx) from implicit equations in mathematica then tell me!)
Thanks!

2. ## Re: Differentiating Implicit Equations

Hello, smatik!

A few SPACES would be welcome.

sin^4(\[Alpha])/x+cos^4(\[Alpha])/y-1/(x+y)==0

You seem to have: . $\sin^4\alpha \div x + \cos^4\alpha \div y - 1 \div(x+y)\:==\: 0$

I see several interpretations for this . . .

. . $\frac{\sin^4\alpha}{x} + \frac{\cos^4\alpha}{y} - \frac{1}{x+y} \:=\:0$

. . $\sin^4\left(\frac{\alpha}{x}\right) + \cos^4\left(\frac{\alpha}{y}\right) - \frac{1}{x+y} \:=\:0$

. . and so on.

3. ## Re: Differentiating Implicit Equations

The first one is it!!!! How to write like you did in this forum??

4. ## Re: Differentiating Implicit Equations

If you hover your mouse over a math expression rendered in latex on this forum you can see the latex code for that expression.

To render the code, surround your code in  tags, for example:

$$\frac{2}{3\cos\alpha}$$ gives:

$\frac{2}{3\cos\alpha}$

5. ## Re: Differentiating Implicit Equations

Thanks.I meant this this.....
$\frac{sin^4(\alpha)}{x}+\frac{cos^4(\alpha)}{x}-\frac{1}{x+y} = 0$

6. ## Re: Differentiating Implicit Equations

About the dy/dx thing. I haven't actually checked, but I believe that the Mathematica uses the form "d/dx(...)" for derivatives. Technically speaking this is probably a partial derivative.

-Dan

7. ## Re: Differentiating Implicit Equations

Hello, smatik!

$\text{Differentiate: }\:\frac{\sin^4\!\alpha}{x}+\frac{\cos^4\!\alpha}{ y}-\frac{1}{x+y} \:=\: 0$

Since $\alpha$ is a constant,
. . let: $a \,=\,\sin^4\!\alpha,\;b \,=\,\cos^4\!\alpha$

We have: . $ax^{\text{-}1} + by^{\text{-}1} - (x+y)^{\text{-}1} \:=\:0$

Then: . $-ax^{\text{-}2} - by^{\text{-}2}y' - (x+y)^{\text{-}2}(1+y') \:=\:0$

. . . . . $-\frac{a}{x^2} - \frac{b}{y^2}y' - \frac{1}{(x+y)^2} - \frac{1}{(x+y)^2}y' \:=\:0$

Multiply by $x^2y^2(x+y)^2\!:$

. . $-ay^2(x+y)^2 - bx^2(x+y)^2y' - x^2y^2 - x^2y^2y' \:=\:0$

. . . . $-bx^2(x+y)^2y' - x^2y^2y' \:=\:ay^2(x+y)^2 + x^2y^2$

. . . . $-x^2\left[b(x+y)^2 + y^2\right]y' \:=\:y^2\left[a(x+y)^2 + x^2\right]$

. . . . . . . . . . . $y' \;=\;-\frac{y^2\left[a(x+y)^2 + x^2\right]}{x^2\left[b(x+y)^2 + y^2\right]}$

. . . . . . . . . . . $y' \;=\;-\frac{y^2\left[(x+y)^2\sin^4\!\alpha + x^2\right]}{x^2\left[(x+y)^2\cos^4\!\alpha + y^2\right]}$