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Answer 1 · 2017-10-16T05:41:10

$\frac{d y}{d x} = \frac{2}{3} {csc}^{2} (\frac{x}{3})$ $dy/dx=(2)/3csc^2(x/3)$ where $csc$ $csc$ is cosec.

$y = - 2 cot (\frac{x}{3})$ $y=-2cot(x/3)$

Differentiate both sides

$\frac{d}{d x} (y) = \frac{d}{d x} (- 2 cot (\frac{x}{3}))$ $d/dx(y)=d/dx(-2cot(x/3))$

Apply the chain rule.

As we know the derivative of $cot x$ $cotx$ is ${csc}^{2} x$ $csc^2x$

$\frac{d y}{d x} = - 2 {csc}^{2} (\frac{x}{3}) \frac{d}{d x} (\frac{x}{3})$ $dy/dx=-2csc^2(x/3)d/dx(x/3)$

$\frac{d y}{d x} = 2 {csc}^{2} (\frac{x}{3}) \cdot \frac{1}{3}$ $dy/dx=2csc^2(x/3)*1/3$

$\frac{d y}{d x} = \frac{2}{3} {csc}^{2} (\frac{x}{3})$ $dy/dx=(2)/3csc^2(x/3)$