Find dz/dy z^2=x^2+y^2

$z^{2} = x^{2} + y^{2}$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = z$ .

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

$\frac{d}{d u} [u^{2}] \frac{d}{d y} [z]$

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

$2 u \frac{d}{d y} [z]$

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

$2 z \frac{d}{d y} [z]$

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

$2 z z'$

Step 3

Differentiate the right side of the equation.

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

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

Since $x^{2}$ is constant with respect to $y$ , the derivative of $x^{2}$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [y^{2}]$

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

$0 + 2 y$

Add $0$ and $2 y$ .

$2 y$

Step 4

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

$2 z z' = 2 y$

Step 5

Divide each term in $2 z z' = 2 y$ by $2 z$ and simplify.

Divide each term in $2 z z' = 2 y$ by $2 z$ .

$\frac{2 z z'}{2 z} = \frac{2 y}{2 z}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 z z'}{2 z} = \frac{2 y}{2 z}$

Rewrite the expression.

$\frac{z z'}{z} = \frac{2 y}{2 z}$

Cancel the common factor of $z$ .

Cancel the common factor.

$\frac{z z'}{z} = \frac{2 y}{2 z}$

Divide $z'$ by $1$ .

$z' = \frac{2 y}{2 z}$

Simplify the right side.

Cancel the common factor of $2$ .

Cancel the common factor.

$z' = \frac{2 y}{2 z}$

Rewrite the expression.

$z' = \frac{y}{z}$

Step 6

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

$\frac{d z}{d y} = \frac{y}{z}$

Find dz/dy z^2=x^2+y^2

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