Easy question: Why is $\dfrac{\partial \sin y(t)}{\partial t} = y'(t)\cos t$ when $\dfrac{\partial\sin y(t)}{\partial y} = 0\text{ ?}$
See: http://www.wolframalpha.com/input/?i=partial+derivative+sin(y(t))+wrt+t for more.
2026-03-28 10:16:31.1774692991
Why is partial $\dfrac{\partial \sin y(t)}{\partial t} = y'(t)\cos t$ when $\dfrac{\partial\sin y(t)}{\partial y} = 0\text{ ?}$
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This is because of the chain rule, when you take the derivative of a function within a function you multiply their derivatives.
Therefore, $\partial_t(\sin(y(t))) = y'(t)\cdot\cos(y(t))$
The reason why $\partial_y(\sin(y(t))) =0$ is because the given function is independent of the variable $y$.
$y(t)$ should not be confused with $y$, one is function notation while the other is a variable.
If you want to calculate $\partial_{y(t)}\sin(y(t))$ this would give ;
$\partial_{y(t)}\sin(y(t)) = \cos(y(t))$