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Find $\frac{dy}{dx}$ of $\tan(2x+y) = 2x$.

All of the possible answer choices do not include a trigonometric function in them, so I figured there might be some substitution required.

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So far, the process I took was to differentiate both sides and use the chain rule for the left side.

$$\sec^2(2x+y)(2 + y') = 2$$ $$2 + y' = \frac{2}{\sec^2(2x+y)}$$ $$y' = \frac{2}{\sec^2(2x+y)} - 2$$ $$y' = \frac{2}{\sec^2(2x+y)} - \frac{2\sec^2(2x+y)}{\sec^2(2x+y)}$$ $$y' = \frac{2(1-\sec^2(2x+y))}{\sec^2(2x+y)}$$ $$y' = \frac{-2\tan^2(2x+y)}{\sec^2(2x+y)}$$ $$y' = -2\sin^2(2x+y)$$

I'm stuck here because I'm not sure if there are any substitutions to make here to remove the trigonometric function from the answer.

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$y' = \frac{2}{\sec^2(2x+y)} - 2$

$y' = \frac{2}{1+\tan^2(2x+y)} - 2$

And now we substitute from the original!

$tan(2x+y) = 2x$

$y' = \frac{2}{1+4x^2} - 2$

$y' = \frac{2-2-8x^2}{1+4x^2}$

$y' = \frac{-8x^2}{1+4x^2}$

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