Using implicit differentiation, find $\frac{d y}{d x}$ . $\newline$ $-7 \cos (3 x) \sin (3 y)=4 x-1$

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Simplify expressions using trigonometric identities

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Q. Using implicit differentiation, find

\frac{d y}{d x}

\newline

-7 \cos (3 x) \sin (3 y)=4 x-1

Given Equation: We are given the equation $-7\cos(3x)\sin(3y)=4x-1$ . To find $\frac{dy}{dx}$ , we will differentiate both sides of the equation with respect to $x$ , using the chain rule for the functions of $y$ , since $y$ is a function of $x$ .
Differentiate Left Side: Differentiate the left side of the equation with respect to $x$ . The derivative of $-7\cos(3x)\sin(3y)$ with respect to $x$ is $-7[-\sin(3x)(3)(\sin(3y)) + \cos(3x)(3)(\cos(3y))(\frac{dy}{dx})]$ by using the product rule and chain rule.
Differentiate Right Side: Differentiate the right side of the equation with respect to $x$ . The derivative of $4x-1$ with respect to $x$ is $4$ , since the derivative of a constant is $0$ .
Simplify Left Side: Now we have the equation $-7[-\sin(3x)(3)(\sin(3y)) + \cos(3x)(3)(\cos(3y))\left(\frac{dy}{dx}\right)] = 4$ . Simplify the left side by distributing the $-7$ and the $3$ inside the brackets.
Isolate $\frac{dy}{dx}$ Term: After simplifying, we get $21\sin(3x)\sin(3y) - 21\cos(3x)\cos(3y)\frac{dy}{dx} = 4$ .
Move Term to Other Side: We want to solve for $\frac{dy}{dx}$ , so we need to isolate the term containing $\frac{dy}{dx}$ . Move the term without $\frac{dy}{dx}$ to the other side of the equation by adding $21\sin(3x)\sin(3y)$ to both sides.
Divide by Constant: We now have $-21\cos(3x)\cos(3y)\left(\frac{dy}{dx}\right) = 4 + 21\sin(3x)\sin(3y)$ .
Final Expression: Divide both sides of the equation by $-21\cos(3x)\cos(3y)$ to solve for $(\frac{dy}{dx})$ .
Final Expression: Divide both sides of the equation by $-21\cos(3x)\cos(3y)$ to solve for $\frac{dy}{dx}$ .The final expression for $\frac{dy}{dx}$ is $\frac{dy}{dx} = \frac{4 + 21\sin(3x)\sin(3y)}{-21\cos(3x)\cos(3y)}$ .