Find dx/dy xsin(y)+ysin(x)=6

$x \sin (y) + y \sin (x) = 6$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x \sin (y) + y \sin (x)) = \frac{d}{d y} (6)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x \sin (y) + y \sin (x)$ with respect to $y$ is $\frac{d}{d y} [x \sin (y)] + \frac{d}{d y} [y \sin (x)]$ .

$\frac{d}{d y} [x \sin (y)] + \frac{d}{d y} [y \sin (x)]$

Evaluate $\frac{d}{d y} [x \sin (y)]$ .

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = x$ and $g (y) = \sin (y)$ .

$x \frac{d}{d y} [\sin (y)] + \sin (y) \frac{d}{d y} [x] + \frac{d}{d y} [y \sin (x)]$

The derivative of $\sin (y)$ with respect to $y$ is $\cos (y)$ .

$x \cos (y) + \sin (y) \frac{d}{d y} [x] + \frac{d}{d y} [y \sin (x)]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$x \cos (y) + \sin (y) x' + \frac{d}{d y} [y \sin (x)]$

Evaluate $\frac{d}{d y} [y \sin (x)]$ .

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = y$ and $g (y) = \sin (x)$ .

$x \cos (y) + \sin (y) x' + y \frac{d}{d y} [\sin (x)] + \sin (x) \frac{d}{d y} [y]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \sin (y)$ and $g (y) = x$ .

To apply the Chain Rule, set $u$ as $x$ .

$x \cos (y) + \sin (y) x' + y (\frac{d}{d u} [\sin (u)] \frac{d}{d y} [x]) + \sin (x) \frac{d}{d y} [y]$

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$x \cos (y) + \sin (y) x' + y (\cos (u) \frac{d}{d y} [x]) + \sin (x) \frac{d}{d y} [y]$

Replace all occurrences of $u$ with $x$ .

$x \cos (y) + \sin (y) x' + y (\cos (x) \frac{d}{d y} [x]) + \sin (x) \frac{d}{d y} [y]$

Rewrite $\frac{d}{d y} [x]$ as $x'$ .

$x \cos (y) + \sin (y) x' + y (\cos (x) x') + \sin (x) \frac{d}{d y} [y]$

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 1$ .

$x \cos (y) + \sin (y) x' + y (\cos (x) x') + \sin (x) \cdot 1$

Multiply $\sin (x)$ by $1$ .

$x \cos (y) + \sin (y) x' + y \cos (x) x' + \sin (x)$

Step 3

Since $6$ is constant with respect to $y$ , the derivative of $6$ with respect to $y$ is $0$ .

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$x \cos (y) + \sin (y) x' + y \cos (x) x' + \sin (x) = 0$

Step 5

Solve for $x'$ .

Simplify the left side.

Reorder factors in $x \cos (y) + \sin (y) x' + y \cos (x) x' + \sin (x)$ .

$x \cos (y) + x' \sin (y) + y x' \cos (x) + \sin (x) = 0$

Move all terms not containing $x'$ to the right side of the equation.

Subtract $x \cos (y)$ from both sides of the equation.

$x' \sin (y) + y x' \cos (x) + \sin (x) = - x \cos (y)$

Subtract $\sin (x)$ from both sides of the equation.

$x' \sin (y) + y x' \cos (x) = - x \cos (y) - \sin (x)$

Factor $x'$ out of $x' \sin (y) + y x' \cos (x)$ .

Factor $x'$ out of $x' \sin (y)$ .

$x' (\sin (y)) + y (x') \cos (x) = - x \cos (y) - \sin (x)$

Factor $x'$ out of $y x' \cos (x)$ .

$x' (\sin (y)) + x' (y \cos (x)) = - x \cos (y) - \sin (x)$

Factor $x'$ out of $x' (\sin (y)) + x' (y \cos (x))$ .

$x' (\sin (y) + y \cos (x)) = - x \cos (y) - \sin (x)$

Divide each term in $x' (\sin (y) + y \cos (x)) = - x \cos (y) - \sin (x)$ by $\sin (y) + y \cos (x)$ and simplify.

Divide each term in $x' (\sin (y) + y \cos (x)) = - x \cos (y) - \sin (x)$ by $\sin (y) + y \cos (x)$ .

$\frac{x' (\sin (y) + y \cos (x))}{\sin (y) + y \cos (x)} = \frac{- x \cos (y)}{\sin (y) + y \cos (x)} + \frac{- \sin (x)}{\sin (y) + y \cos (x)}$

Simplify the left side.

Cancel the common factor of $\sin (y) + y \cos (x)$ .

Cancel the common factor.

$\frac{x' (\sin (y) + y \cos (x))}{\sin (y) + y \cos (x)} = \frac{- x \cos (y)}{\sin (y) + y \cos (x)} + \frac{- \sin (x)}{\sin (y) + y \cos (x)}$

Divide $x'$ by $1$ .

$x' = \frac{- x \cos (y)}{\sin (y) + y \cos (x)} + \frac{- \sin (x)}{\sin (y) + y \cos (x)}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$x' = - \frac{x \cos (y)}{\sin (y) + y \cos (x)} + \frac{- \sin (x)}{\sin (y) + y \cos (x)}$

Move the negative in front of the fraction.

$x' = - \frac{x \cos (y)}{\sin (y) + y \cos (x)} - \frac{\sin (x)}{\sin (y) + y \cos (x)}$

Simplify terms.

Combine the numerators over the common denominator.

$x' = \frac{- x \cos (y) - \sin (x)}{\sin (y) + y \cos (x)}$

Factor $- 1$ out of $- x \cos (y)$ .

$x' = \frac{- (x \cos (y)) - \sin (x)}{\sin (y) + y \cos (x)}$

Factor $- 1$ out of $- \sin (x)$ .

$x' = \frac{- (x \cos (y)) - (\sin (x))}{\sin (y) + y \cos (x)}$

Factor $- 1$ out of $- (x \cos (y)) - (\sin (x))$ .

$x' = \frac{- (x \cos (y) + \sin (x))}{\sin (y) + y \cos (x)}$

Simplify the expression.

Rewrite $- (x \cos (y) + \sin (x))$ as $- 1 (x \cos (y) + \sin (x))$ .

$x' = \frac{- 1 (x \cos (y) + \sin (x))}{\sin (y) + y \cos (x)}$

Move the negative in front of the fraction.

$x' = - \frac{x \cos (y) + \sin (x)}{\sin (y) + y \cos (x)}$

Step 6

Replace $x'$ with $\frac{d x}{d y}$ .

$\frac{d x}{d y} = - \frac{x \cos (y) + \sin (x)}{\sin (y) + y \cos (x)}$

Find dx/dy xsin(y)+ysin(x)=6

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