If $x = y z$ $x = yz$ and $y = 2 sin (y + z)$ $y = 2 sin(y + z)$ , find $\frac{d x}{d y}$ $(dx)/(dy)$ ?

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If $x = y z$ $x = yz$ and $y = 2 sin (y + z)$ $y = 2 sin(y + z)$ , find $\frac{d x}{d y}$ $(dx)/(dy)$ ?

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Answer 1 · 2017-12-16T16:44:37

$\frac{d x}{d y} = arcsin (\frac{1}{2} y) + \frac{y}{2 \sqrt{1 - {(\frac{1}{2} y)}^{2}}} - 2 y$ $dx/dy = arcsin(1/2y) + y/(2sqrt(1 - (1/2y)^2)) - 2y$

Explanation:

From our first equation, we know that $z = \frac{x}{y}$ $z = x/y$ .

Therefore,

$y = 2 sin (y + \frac{x}{y})$ $y =2sin(y + x/y)$

Solving for $x$ $x$ will prove quite easy (since we are differentiating with respect to $x$ $x$ , this will probably make some sense).

$\frac{1}{2} y = sin (y + \frac{x}{y})$ $1/2y = sin(y + x/y)$

$arcsin (\frac{1}{2} y) = y + \frac{x}{y}$ $arcsin(1/2y) = y + x/y$

$arcsin (\frac{1}{2} y) - y = \frac{x}{y}$ $arcsin(1/2y) - y = x/y$

$y arcsin (\frac{1}{2} y) - y^{2} = x$ $yarcsin(1/2y) - y^2 = x$

Now we differentiate this and we will get our answer.

For the first term, $y arcsin (\frac{1}{2} y)$ $yarcsin(1/2y)$ , we use the chain and product rules. We know the derivative of $arcsin y$ $arcsiny$ to be $\frac{1}{\sqrt{1 - y^{2}}}$ $1/sqrt(1 - y^2)$ , thus the derivative of $arcsin (\frac{1}{2} y) = \frac{1}{2 \sqrt{1 - {(\frac{1}{2} y)}^{2}}}$ $arcsin(1/2y) = 1/(2sqrt(1 - (1/2y)^2))$ . By the product rule,

$arcsin (\frac{1}{2} y) + \frac{y}{2 \sqrt{1 - {(\frac{1}{2} y)}^{2}}} - 2 y = \frac{d x}{d y}$ $arcsin(1/2y) + y/(2sqrt(1 -(1/2y)^2)) - 2y = (dx)/(dy)$

Hopefully this helps!

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If $x = y z$ $x = yz$ and $y = 2 sin (y + z)$ $y = 2 sin(y + z)$ , find $\frac{d x}{d y}$ $(dx)/(dy)$ ?

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Explanation:

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