Questions: Find dy/dx by implicit differentiation. tan^(-1)(3x^2y) = x + 4xy^2 dy/dx =

Solution Steps

To find \(\frac{d y}{d x}\) by implicit differentiation, follow these steps:

1. Differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\).
2. Apply the chain rule where necessary.
3. Collect all terms involving \(\frac{d y}{d x}\) on one side of the equation.
4. Solve for \(\frac{d y}{d x}\).

Given the equation: \[ \tan^{-1}(3x^2 y) = x + 4xy^2 \]

We need to find \(\frac{d y}{d x}\) by implicit differentiation. Start by differentiating both sides with respect to \( x \).

For the left side, use the chain rule: \[ \frac{d}{dx} \left( \tan^{-1}(3x^2 y) \right) = \frac{1}{1 + (3x^2 y)^2} \cdot \frac{d}{dx} (3x^2 y) \]

Now, differentiate \(3x^2 y\) with respect to \( x \): \[ \frac{d}{dx} (3x^2 y) = 3x^2 \frac{d y}{d x} + 6xy \]

So, the left side becomes: \[ \frac{1}{1 + (3x^2 y)^2} \cdot (3x^2 \frac{d y}{d x} + 6xy) \]

For the right side, differentiate term by term: \[ \frac{d}{dx} (x + 4xy^2) = 1 + 4y^2 + 8xy \frac{d y}{d x} \]

Equate the derivatives from both sides: \[ \frac{1}{1 + (3x^2 y)^2} \cdot (3x^2 \frac{d y}{d x} + 6xy) = 1 + 4y^2 + 8xy \frac{d y}{d x} \]

Multiply both sides by \(1 + (3x^2 y)^2\) to clear the fraction: \[ 3x^2 \frac{d y}{d x} + 6xy = (1 + (3x^2 y)^2)(1 + 4y^2 + 8xy \frac{d y}{d x}) \]

Distribute on the right side: \[ 3x^2 \frac{d y}{d x} + 6xy = 1 + 4y^2 + (3x^2 y)^2 + 8xy \frac{d y}{d x} + 4y^2 (3x^2 y)^2 \]

Combine like terms involving \(\frac{d y}{d x}\): \[ 3x^2 \frac{d y}{d x} - 8xy \frac{d y}{d x} = 1 + 4y^2 + (3x^2 y)^2 + 4y^2 (3x^2 y)^2 - 6xy \]

Factor out \(\frac{d y}{d x}\): \[ (3x^2 - 8xy) \frac{d y}{d x} = 1 + 4y^2 + (3x^2 y)^2 + 4y^2 (3x^2 y)^2 - 6xy \]

Finally, solve for \(\frac{d y}{d x}\): \[ \frac{d y}{d x} = \frac{1 + 4y^2 + (3x^2 y)^2 + 4y^2 (3x^2 y)^2 - 6xy}{3x^2 - 8xy} \]

Final Answer