2016年2月15日 · If, say, $-\pi/2 \le x \le \pi/2$, and $-\pi/2 \le y \le \pi/2$, then $$\sin(xy) = \sin(\arcsin(\sin(x)) \arcsin(\sin(y)))$$ expresses $\sin(xy)$ as a function of ...

2015年6月17日 · No rearrange this equation as #dy/dx-xcos(xy) dy/dx = ycos(xy)#, factor out the #dy/dx# on the left-hand side and then divide both sides by #1-xcos(xy)# to get #dy/dx=\frac{ycos(xy)}{1-xcos(xy)}# Since the original equation cannot be solved explicitly for #y# as a function of #x# , this is the best you can do.

2016年10月17日 · See all questions in Intuitive Approach to the derivative of y=sin(x) Impact of this question 44617 views around the world

Me and my mates are crunching this question for a while now. While we know that $\sin(xy)$ is continuous ...

2018年7月19日 · I think the other answers don't really address the heart of the question, and are too focused on proving why @Ethan_Chan's equations are false specifically for trigonometric functions.

2015年3月27日 · first you have to know that $$\lim_{x\to 0}\dfrac{\sin x}{x} = \lim_{x\to 0}\dfrac{\cos x}{1} = 1$$ (I used L'Hopital)

2015年7月22日 · $$\int \sin xy \ dy = \int \frac{\sin u}{x} du$$ $$ \frac{-\cos u}{x} + c = \frac{-\cos xy}{x} + c$$ Hint: when integrating wrt y, x becomes a constant so you don't need to worry about it ($\sin xy $ is a function of two variables).

2018年10月26日 · Exactly the same reason with $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ shows that $$\lim\limits_{(x,y)\to(0,0)}\frac{\sin xy}{xy}=1$$ with the common formula $$\cos xy\leq\dfrac{\sin xy}{xy}\leq1$$ we can conclude the lmit is $1$ as the limit in one variable shows.

$\begingroup$ we can write it as sin(x)-sin(y)-sin(x*y)=0 and use some numerical approximation method newton's method or something like this. first trivial solution is x=y=0 $\endgroup$ – dato datuashvili

2016年11月10日 · Start by expanding sin(x + y) using the sum and difference identity. sin(x + y) = xy sinxcosy + cosxsiny = xy By the product rule: cosxcosy + sinx xx -siny(dy/dx ...