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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.

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  • $\begingroup$ how did you calculate that? because that's incorrect $\endgroup$ – peek-a-boo Jul 20 at 16:57
  • $\begingroup$ @peek-a-boo I checked it on a website called symbolab as well and there I get -y/x $\endgroup$ – muhammad haider Jul 20 at 17:03
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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 $\cos xy$.

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  • $\begingroup$ the formula I used was dy/dx= - (derivative with x as the variable) / (derivative with y as the variable). I also checked the answer on the website symbolab $\endgroup$ – muhammad haider Jul 20 at 17:05
  • $\begingroup$ If you want to do that, then you need to set up the equation $F(x,y)=0.$ Then, you'll find that $y'=-F_x/F_y.$ In this case, $F(x,y)=\cos xy-y,$ $F_x=-y\sin (xy) ,$ and $F_y=-x\sin (xy)-1.$ Putting this together gives the same as my answer. $\endgroup$ – cmk Jul 20 at 17:10
  • $\begingroup$ You must've entered it wrong. See symbolab.com/solver/implicit-derivative-calculator/… $\endgroup$ – cmk Jul 20 at 17:11
  • $\begingroup$ Technically, you only need $F(x,y)=c$, for some constant $c$, but $c=0$ here. $\endgroup$ – cmk Jul 20 at 17:16
  • $\begingroup$ i see. I forgot to bring y to the other side, I understand. And to find it at x=0 I just plug 0 in and 0 would be the answer $\endgroup$ – muhammad haider Jul 20 at 17:16
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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.

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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 $$\frac{dy}{dx}=-\frac{\frac{\partial F(x,y)}{\partial x} } {\frac{\partial F(x,y)}{\partial y} }=-\frac{y \sin(xy) }{1+x \sin(y) }$$

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