Given the equation $$ 9 \sin 5 x y=1 $$, evaluate $$ \frac{d y}{d x} $$. Assume that the equation implicitly defines $$ y $$ as a differentiable function of $$ x $$. If $$ F(x, y)=9 \sin 5 x y-1 $$, then $$ F_{x}=45 y \cos (5 x y) $$. If $$ x, y)=9 \sin 5 x y-1 $$, then $$ F_{y}= $$

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We start with

$$
F(x, y) = 9 \sin (5xy) - 1.
$$

To compute $$ F_{y} $$, we differentiate $$ F(x, y) $$ with respect to $$ y $$ holding $$ x $$ constant.

The derivative of $$ \sin (5xy) $$ with respect to $$ y $$ is found using the chain rule. First, differentiate the outer function:

$$
\frac{d}{d y} \sin (5xy) = \cos (5xy) \cdot \frac{d}{d y}(5xy).
$$

Next, differentiate the inner function $$ 5xy $$ with respect to $$ y $$:

$$
\frac{d}{dy}(5xy) = 5x.
$$

Thus,

$$
\frac{d}{d y} \sin (5xy) = \cos (5xy) \cdot 5x = 5x \cos (5xy).
$$

Multiplying by the constant coefficient 9 from the original function, we have

$$
F_y = 9 \cdot 5x \cos (5xy) = 45x \cos (5xy).
$$
$$ F_y = 45x \cos (5xy) $$.

Given the equation \( 9 \sin 5 x y=1 \), evaluate \( \frac{d y}{d x} \). Assume that the equation implicitly defines \( y \) as a differentiable function of \( x \). If \( F(x, y)=9 \sin 5 x y-1 \), then \( F_{x}=45 y \cos (5 x y) \). If \( x, y)=9 \sin 5 x y-1 \), then \( F_{y}= \)

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