1. Differentiation

I'm not sure with this particular problem, do I use implicit differentiation or just normal differentiation?

x^2 + y^2 = arcsin(xy)

Can I assume that it =1? => x^2 + y^2 = 1 (unit circle)

If so, how do I find dy/dx of 1 = arcsin(xy)?

From an attempt, does dy/dx = ± 1/x ?

Any help is much appreciated,

Dranalion

2. Originally Posted by Dranalion
I'm not sure with this particular problem, do I use implicit differentiation or just normal differentiation?

x^2 + y^2 = arcsin(xy)

Can I assume that it =1? => x^2 + y^2 = 1 (unit circle)

If so, how do I find dy/dx of 1 = arcsin(xy)?

Any help is much appreciated,

Dranalion
No you can't.

You need to use implicit differentiation.

The derivative of the left hand side is $\displaystyle 2x + 2y \cdot \frac{dy}{dx}$. Your first task now is to differentiate the right hand side with respect to x, treating y as an implicit function of x.

3. For the right hand side, I know that d/dx(arcsin(x)) = 1/sqrt(1 - x^2)

So, for this; d/dx(arcsin(xy)) = 1/sqrt(1-[xy]^2)

Is this correct? If so, where to now? I am unsure

4. Originally Posted by Dranalion
For the right hand side, I know that d/dx(arcsin(x)) = 1/sqrt(1 - x^2)

So, for this; d/dx(arcsin(xy)) = 1/sqrt(1-[xy]^2)

Is this correct? If so, where to now? I am unsure
Since xy is a function of x, you have to multiply the above result by the derivative with respect to x of xy (this follows from the chain rule). To (implicitly) differentiate xy, use the product rule.

5. Sorry, I am a little confused..

Is it possible to go d/dx(arcsin(xy)) = y / sqrt[1-(xy)^2]

6. Originally Posted by mr fantastic
Since xy is a function of x, you have to multiply the above result by the derivative with respect to x of xy (this follows from the chain rule). To (implicitly) differentiate xy, use the quotient rule.
What do you mean by "To (implicitly) differentiate xy, use the quotient rule."?

Do you mean use the quotient rule for y/sqrt[1-(xy)^2]? If so, hasn't it already been derived?

7. I can only think that mr fantastic mean "to differentiate xy, use the product rule". (xy)'= x'y+ xy' and, since we are differentiating with respect to x, x'= 1, we have (xy)'= y+ xy'.

8. Does this mean to use the chain rule? What happens to the arcsin component?

Perhaps arcsin(u), where u = xy?
du/dx = y + xy' ?
dy/du = 1/sqrt[1 - u^2]
dy/dx = 1/sqrt[1 - (xy)^2] x y + xy'

This is one complicated and lengthy question for me! :P

9. The derivative of $\displaystyle x^2+ y^2$ with respect to x is 2x+ 2y dy/dx. The derivative of arcsin(xy) is a little harder.

Presumably, you know that the derivative of sin(u) with respect to u is cos(u). Since here we have arcsine, write y= arcsin(x) and, from that, x= sin(y). Now, differentiate with respect to y: dx/dy= cos(y). From that, dy/dx= 1/cos(y) (that is NOT just "inverting" both sides but it can be shown to work that way). But now we want that in terms of x. Since sine is "opposite side divided by hypotenuse" in the simple right triangle case, we can imagine "sin(y)= x" as referring to a right triangle with angle y, "opposite side" x, and "hypotenuse" 1. Using the Pythagorean theorem, the "near side" must be $\displaystyle \sqrt{1- x^2}$. Since cosine is "near side divide by hypotenuse", we have $\displaystyle cos(y)= \sqrt{1- x^2}/1= \sqrt{1- x^2}$ and so

$\displaystyle d(arcsin(x))/dx= dy/dx= \frac{1}{cos(y)}= \frac{1}{\sqrt{1- x^2}}$.
a formula you probably have seen in your textbook.

Now use the chain rule. The derivative of arcsin(xy), with respect to x, is $\displaystyle \frac{1}{\sqrt{1- (xy)^2}}$ times the derivative of xy with respect to x: x dy/dx+ y. Putting those together,
$\displaystyle d(arcsin(xy))/dx= \frac{x dy/dx+ y}{\sqrt{1- x^2y^2}}$
so
$\displaystyle 2x+ 2y dy/dx= \frac{x dy/dx+ y}{\sqrt{1- x^2y^2}}$

Solve that for dy/dx.

10. Out of interest, how do you solve for dy/dx if it is in that position?

11. The equation is $\displaystyle 2x+ 2y dy/dx= \frac{x dy/dx+ y}{\sqrt{1- x^2y^2}}$ which is linear in dy/dx. (Because differentiation itself is a linear operator, implicit differentiation always gives equations that are linear in the derivative.)

Just to make the writing easier, I am going to let $\displaystyle \sqrt{1- x^2y^2}$ be "A". $\displaystyle 2x+ 2y\frac{dy}{dx}= \frac{x}{A}\frac{dy}{dx}+ \frac{y}{A}$. Then $\displaystyle 2y\frac{dy}{dx}- \frac{x}{A}\frac{dy}{dx}= \frac{y}{A}- 2x$.

$\displaystyle (2y- \frac{x}{A})\frac{dy}{dx}= \frac{y}{A}- 2x$
$\displaystyle \frac{dy}{dx}= \frac{\frac{y}{A}- 2x}{2y- \frac{x}{A}}$

You can clean that up a little by multiplying both numerator and denominator by A:
$\displaystyle \frac{dy}{dx}= \frac{y- 2Ax}{2Ay- x}= \frac{y- 2x\sqrt{1- x^2y^2}}{2y\sqrt{1- x^2y^2}- x}$