Find dx/dy xsin(x)+y^2=4

$x \sin (x) + y^{2} = 4$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x \sin (x) + y^{2}) = \frac{d}{d y} (4)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x \sin (x) + y^{2}$ with respect to $y$ is $\frac{d}{d y} [x \sin (x)] + \frac{d}{d y} [y^{2}]$ .

$\frac{d}{d y} [x \sin (x)] + \frac{d}{d y} [y^{2}]$

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

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 $f (y) = x$ and $g (y) = \sin (x)$ .

$x \frac{d}{d y} [\sin (x)] + \sin (x) \frac{d}{d y} [x] + \frac{d}{d y} [y^{2}]$

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 $x$ .

$x (\frac{d}{d u} [\sin (u)] \frac{d}{d y} [x]) + \sin (x) \frac{d}{d y} [x] + \frac{d}{d y} [y^{2}]$

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

$x (\cos (u) \frac{d}{d y} [x]) + \sin (x) \frac{d}{d y} [x] + \frac{d}{d y} [y^{2}]$

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

$x (\cos (x) \frac{d}{d y} [x]) + \sin (x) \frac{d}{d y} [x] + \frac{d}{d y} [y^{2}]$

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

$x (\cos (x) x') + \sin (x) \frac{d}{d y} [x] + \frac{d}{d y} [y^{2}]$

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

$x (\cos (x) x') + \sin (x) x' + \frac{d}{d y} [y^{2}]$

Differentiate using the Power Rule.

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 2$ .

$x (\cos (x) x') + \sin (x) x' + 2 y$

Reorder terms.

$x \cos (x) x' + \sin (x) x' + 2 y$

Step 3

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

$0$

Step 4

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

$x \cos (x) x' + \sin (x) x' + 2 y = 0$

Step 5

Solve for $x'$ .

Simplify the left side.

Reorder factors in $x \cos (x) x' + \sin (x) x' + 2 y$ .

$x x' \cos (x) + x' \sin (x) + 2 y = 0$

Subtract $2 y$ from both sides of the equation.

$x x' \cos (x) + x' \sin (x) = - 2 y$

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

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

$x' (x \cos (x)) + x' (\sin (x)) = - 2 y$

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

$x' (x \cos (x)) + x' (\sin (x)) = - 2 y$

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

$x' (x \cos (x) + \sin (x)) = - 2 y$

Divide each term in $x' (x \cos (x) + \sin (x)) = - 2 y$ by $x \cos (x) + \sin (x)$ and simplify.

Divide each term in $x' (x \cos (x) + \sin (x)) = - 2 y$ by $x \cos (x) + \sin (x)$ .

$\frac{x' (x \cos (x) + \sin (x))}{x \cos (x) + \sin (x)} = \frac{- 2 y}{x \cos (x) + \sin (x)}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x' (x \cos (x) + \sin (x))}{x \cos (x) + \sin (x)} = \frac{- 2 y}{x \cos (x) + \sin (x)}$

Divide $x'$ by $1$ .

$x' = \frac{- 2 y}{x \cos (x) + \sin (x)}$

Simplify the right side.

Move the negative in front of the fraction.

$x' = - \frac{2 y}{x \cos (x) + \sin (x)}$

Step 6

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

$\frac{d x}{d y} = - \frac{2 y}{x \cos (x) + \sin (x)}$

Find dx/dy xsin(x)+y^2=4

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