Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

The derivative of $z$ with respect to $x$ is $z'$ .

$z'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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Step 3.1.1

To apply the Chain Rule, set $u$ as $x^{2} + y^{2}$ .

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

Step 3.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [x^{2} + y^{2}]$

Step 3.1.3

Replace all occurrences of $u$ with $x^{2} + y^{2}$ .

$\frac{1}{x^{2} + y^{2}} \frac{d}{d x} [x^{2} + y^{2}]$

Step 3.2

Differentiate.

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Step 3.2.1

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

$\frac{1}{x^{2} + y^{2}} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}])$

Step 3.2.2

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

$\frac{1}{x^{2} + y^{2}} (2 x + \frac{d}{d x} [y^{2}])$

Step 3.2.3

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

$\frac{1}{x^{2} + y^{2}} (2 x + 0)$

Step 3.2.4

Combine fractions.

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Step 3.2.4.1

Add $2 x$ and $0$ .

$\frac{1}{x^{2} + y^{2}} (2 x)$

Step 3.2.4.2

Combine $2$ and $\frac{1}{x^{2} + y^{2}}$ .

$\frac{2}{x^{2} + y^{2}} x$

Step 3.2.4.3

Combine $\frac{2}{x^{2} + y^{2}}$ and $x$ .

$\frac{2 x}{x^{2} + y^{2}}$

Step 4

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

$z' = \frac{2 x}{x^{2} + y^{2}}$

Step 5

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

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