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
The derivative of 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 . Since cosine is "near side divide by hypotenuse", we have and so
.
a formula you probably have seen in your textbook.
Now use the chain rule. The derivative of arcsin(xy), with respect to x, is times the derivative of xy with respect to x: x dy/dx+ y. Putting those together,
so
Solve that for dy/dx.
The equation is 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 be "A". . Then .
You can clean that up a little by multiplying both numerator and denominator by A: