If I differentiate $x\sin(y)$ with respect to $x$ normally then what happens to the $\sin(y)$? In partial differentiation it's taken as constant thus, partially differentiating with respect to $x$ would give the result as $\sin( y)$. I want to know what happens when you differentiate normally e.g. $\frac{dy}{dx}$?
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$\begingroup$Assuming $y$ is a function of $x$,
$$\frac{dy}{dx} = \sin y + x\cos y \frac{dy}{dx} $$
In other words, you're just applying the product rule (and chain rule since $y$ is a function of $x$): Assume $v$ is a function of $g$ which is a function of $x$
$$\frac{d}{dx}(u(x)v(g(x))) = v(g(x))\frac{du}{dx} + u(x)\frac{dv}{dg}\cdot\frac{dg}{dx}$$
$\endgroup$ 10 $\begingroup$You can take $y$ as a function of $x$ and use the chain rule and product rule. $(x\sin (y(x))'=x(\sin(y(x))'+x'\sin(y(x))=xy'(x)\cos(y(x))+\sin(y(x))$
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