Derivative of an implicit Function at a certain point

Question

I wanted to know what the derivative of the function:

$$y=\cos(xy)$$ when $x=0$ was

For the deritive I got $$\frac{dy}{dx}=-\frac{y}{x}$$

so at $x=0$ the derivative of the function is undefined? I wasn't sure if the answer was right so I wanted to check it here.

Answer 1

You should get that $$\frac{dy}{dx}=-\sin (xy)\left(y+x\frac{dy}{dx}\right),$$ or $$\frac{dy}{dx}=-\frac{y\sin xy}{1+x\sin xy}.$$ You must've made a mistake in your implicit differentiation. You might need to think a bit more carefully about how to properly implicitly differentiate $\cos xy$.

Answer 2

the answer may be easily obtained using the general relation: $$ \frac{d}{dx} = \frac{\partial}{\partial x} + \frac{dy}{dx}\frac{\partial}{\partial y} $$

since it is easy to feel a bit confused by the apparent abstraction of this formula, you may check by working from "first principles". then the calculation is straightforward using infinitesimals of the first order. in this framework, we have $\cos(dt) = 1, \sin dt = dt$ for any infinitesimal $dt$, and $dsdt = 0$

using the standard formula $\cos(a+b) = \cos a \cos b - \sin a \sin b$ we get: $$ \begin{align} y+dy &= \cos\bigg(x+dx)(y+dy)\bigg) \\ &= \cos(xy+ydx + xdy) \\ &= y-\sin xy \sin(ydx +xdy) \\ \end{align} $$

this gives: $$ \begin{align} dy = -(ydx + xdy)\sin xy \end{align} $$

from which the result follows on "dividing" through by $dx$, and rearranging.

Answer 3

Fot this kind of problems, I think that defining first the implicit function makes life easier.

For your case, consider $$F(x,y)=y-\cos(xy)=0$$ and compute the partial derivatives $$\frac{\partial F(x,y)}{\partial x}=y \sin(xy)\qquad \qquad \qquad \frac{\partial F(x,y)}{\partial y}=1+x \sin(y)$$ Now, using the implicit function theorem $$\frac{dy}{dx}=-\frac{\frac{\partial F(x,y)}{\partial x} } {\frac{\partial F(x,y)}{\partial y} }=-\frac{y \sin(xy) }{1+x \sin(y) }$$