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Using implicit differentiation, find
(dy)/(dx).

-2cos(3x)sin(3y)=-6x-6

Using implicit differentiation, find $\frac{d y}{d x}$ . $\newline$ $-2 \cos (3 x) \sin (3 y)=-6 x-6$

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Find derivatives of radical functions

Full solution

Q. Using implicit differentiation, find

\frac{d y}{d x}

.

\newline

-2 \cos (3 x) \sin (3 y)=-6 x-6

Identify Equation: Identify the given equation and prepare to use implicit differentiation. $\newline$ Given equation: $-2\cos(3x)\sin(3y)=-6x-6$ $\newline$ Implicit differentiation will be used because $y$ is a function of $x$ , and we need to find $\frac{dy}{dx}.$
Differentiate Both Sides: Differentiate both sides of the equation with respect to $x$ . The left side of the equation requires the product rule and chain rule, while the right side is straightforward. Differentiate the left side: $-2[\cos(3x) * \frac{d}{dx}(\sin(3y)) + \sin(3y) * \frac{d}{dx}(\cos(3x))]$ Differentiate the right side: $\frac{d}{dx}(-6x - 6)$
Apply Chain Rule: Apply the chain rule to the derivatives of the trigonometric functions. $\newline$ For $\sin(3y)$ , the derivative is $\cos(3y) \cdot \frac{d}{dx}(3y) = 3\cos(3y) \cdot \frac{dy}{dx}$ . $\newline$ For $\cos(3x)$ , the derivative is $-\sin(3x) \cdot \frac{d}{dx}(3x) = -3\sin(3x)$ . $\newline$ Now substitute these derivatives into the differentiated left side: $-2[\cos(3x) \cdot (3\cos(3y) \cdot \frac{dy}{dx}) - 3\sin(3x) \cdot \sin(3y)]$ . $\newline$ The right side becomes $-6$ .
Simplify Differentiated Equation: Simplify the differentiated equation. $\newline$ $-2[3\cos(3x)\cos(3y) \cdot \frac{dy}{dx} - 3\sin(3x)\sin(3y)] = -6$ $\newline$ Distribute the $-2$ : $-6\cos(3x)\cos(3y) \cdot \frac{dy}{dx} + 6\sin(3x)\sin(3y) = -6$
Isolate $\frac{dy}{dx}$ : Isolate the term with $\frac{dy}{dx}$ on one side of the equation. $\newline$ $-6\cos(3x)\cos(3y) \cdot \frac{dy}{dx} = -6 - 6\sin(3x)\sin(3y)$
Solve for $\frac{dy}{dx}$ : Solve for $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{-6 - 6\sin(3x)\sin(3y)}{-6\cos(3x)\cos(3y)}$
Simplify the equation by dividing both numerator and denominator by $-6$ .
$\frac{dy}{dx} = \frac{1 + \sin(3x)\sin(3y)}{\cos(3x)\cos(3y)}$

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