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\( \begin{array}{r}\text { Find } d y / d x \text { by implicit differentiation. } \\ \tan ^{-1}\left(3 x^{2} y\right)=x+2 x y^{2} \\ y^{\prime}=\square\end{array} \)

Ask by Schneider Weaver. in the United States
Mar 16,2025

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\( y' = \frac{(1 + 9x^4y^2)(1 + 2y^2) - 6xy}{3x^2 - 4xy(1 + 9x^4y^2)} \)

Solución

We start with the equation \[ \tan^{-1}(3x^2y) = x + 2xy^2. \] Differentiate both sides with respect to \( x \). 1. Differentiate the left-hand side using the chain rule. Recall that \[ \frac{d}{dx}\tan^{-1}(u) = \frac{1}{1+u^2}\frac{du}{dx}. \] Here, \( u = 3x^2y \). Its derivative is \[ \frac{d}{dx}(3x^2y) = 6xy + 3x^2y', \] so the derivative of the left-hand side becomes \[ \frac{6xy+3x^2y'}{1+(3x^2y)^2} = \frac{6xy+3x^2y'}{1+9x^4y^2}. \] 2. Differentiate the right-hand side. The derivative of \( x \) is \( 1 \) and for \( 2xy^2 \) we use the product rule: \[ \frac{d}{dx}(2xy^2) = 2y^2 + 4xyy'. \] Thus, the derivative of the right-hand side is \[ 1 + 2y^2 + 4xyy'. \] 3. Set the derivatives equal: \[ \frac{6xy+3x^2y'}{1+9x^4y^2} = 1 + 2y^2 + 4xyy'. \] 4. Multiply both sides by \( 1+9x^4y^2 \) to eliminate the denominator: \[ 6xy + 3x^2y' = \left(1+9x^4y^2\right)(1+2y^2 + 4xyy'). \] 5. Expand the right-hand side: \[ 6xy + 3x^2y' = \left(1+9x^4y^2\right)(1+2y^2) + 4xyy'\left(1+9x^4y^2\right). \] 6. Gather all terms with \( y' \) on one side: \[ 3x^2y' - 4xyy'\left(1+9x^4y^2\right) = \left(1+9x^4y^2\right)(1+2y^2) - 6xy. \] 7. Factor out \( y' \) on the left-hand side: \[ y'\left(3x^2 - 4xy\left(1+9x^4y^2\right)\right) = \left(1+9x^4y^2\right)(1+2y^2) - 6xy. \] 8. Finally, solve for \( y' \): \[ y' = \frac{\left(1+9x^4y^2\right)(1+2y^2)-6xy}{3x^2-4xy\left(1+9x^4y^2\right)}. \] The final answer is \[ y'=\frac{\left(1+9x^4y^2\right)(1+2y^2)-6xy}{3x^2-4xy\left(1+9x^4y^2\right)}. \]

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Beyond the Answer

To find \( \frac{dy}{dx} \) using implicit differentiation, start by differentiating both sides of the equation \( \tan^{-1}(3x^2y) = x + 2xy^2 \): 1. Differentiate the left side using the chain rule: \[ \frac{d}{dx}\left(\tan^{-1}(3x^2y)\right) = \frac{1}{1 + (3x^2y)^2} \cdot \frac{d}{dx}(3x^2y) = \frac{1}{1 + 9x^4y^2}\left(3(2xy + x^2\frac{dy}{dx})\right) = \frac{6xy + 3x^2y'}{1 + 9x^4y^2} \] 2. Differentiate the right side: \[ \frac{d}{dx}(x + 2xy^2) = 1 + 2y^2 + 2x(2y\frac{dy}{dx}) = 1 + 2y^2 + 4xy\frac{dy}{dx} \] Setting both derivatives equal gives: \[ \frac{6xy + 3x^2y'}{1 + 9x^4y^2} = 1 + 2y^2 + 4xy\frac{dy}{dx} \] Now, clear the fraction by multiplying through by the denominator: \[ 6xy + 3x^2y' = (1 + 2y^2 + 4xy\frac{dy}{dx})(1 + 9x^4y^2) \] This will yield an equation that can be rearranged to solve for \( y' \). After some algebraic manipulation, isolate \( y' \) terms on one side and factor out \( y' \): \[ y'(3x^2 - 4xy(1 + 9x^4y^2)) = \text{other terms} \] Thus, we can express \( y' \) as: \[ y' = \frac{\text{other terms}}{3x^2 - 4xy(1 + 9x^4y^2)} \] So, the value of \( \frac{dy}{dx} \) will depend on the simplified other terms from the right-hand side equation substitution. Pitching the right combination can lead you to the final value to fill in the square!

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