Using implicit differentiation, find $\frac{d y}{d x}$ . $\newline$ $-3 \cos (3 x) \sin (2 y)=5-x$

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Algebra 2

Simplify expressions using trigonometric identities

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Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

-3 \cos (3 x) \sin (2 y)=5-x

Differentiate with product rule: Differentiate both sides of the equation with respect to $x$ . We need to use the product rule for differentiation on the left side, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. We also need to use the chain rule for differentiating composite functions.
Differentiate trigonometric function: Differentiate $-3\cos(3x)\sin(2y)$ with respect to $x$ . Using the product rule, we get: $\frac{d}{dx}[-3\cos(3x)\sin(2y)] = -3[-\sin(3x)\cdot 3\cdot\frac{dy}{dx}\sin(2y) + \cos(3x)\cdot 2\cdot\cos(2y)\cdot\frac{dy}{dx}]$
Differentiate constant term: Differentiate $5-x$ with respect to $x$ . The derivative of $5$ is $0$ , and the derivative of $-x$ is $-1$ , so we get: $\frac{d}{dx}[5-x] = -1$
Set derivatives equal: Set the derivatives from Step $2$ and Step $3$ equal to each other. $\newline$ $-3[-\sin(3x)\cdot 3\cdot\frac{dy}{dx}\sin(2y) + \cos(3x)\cdot 2\cdot\cos(2y)\frac{dy}{dx}] = -1$
Simplify equation: Simplify the equation from Step $4$ . $\newline$ $-3[-3\sin(3x)\sin(2y)\frac{dy}{dx} + 2\cos(3x)\cos(2y)\frac{dy}{dx}] = -1$
Factor out $\frac{dy}{dx}$ : Factor out $\frac{dy}{dx}$ from the left side of the equation. $\newline$ $\frac{dy}{dx}[-3(-3\sin(3x)\sin(2y) + 2\cos(3x)\cos(2y))] = -1$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ . $\frac{dy}{dx} = \frac{-1}{[-3(-3\sin(3x)\sin(2y) + 2\cos(3x)\cos(2y))]}$
Simplify denominator: Simplify the denominator. $\frac{dy}{dx} = \frac{-1}{[9\sin(3x)\sin(2y) - 6\cos(3x)\cos(2y)]}$
Check for errors: Check for any possible simplification or errors. The expression seems to be simplified correctly, and there are no apparent math errors.