Calculus Examples

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

Step 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 $g (x) = x^{2} + y^{2} + 1$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $x^{2} + y^{2} + 1$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u$ with $x^{2} + y^{2} + 1$ .

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

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $x^{2} + y^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [1]$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Add $2 x$ and $0$ .

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

Step 2.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 2.6

Combine fractions.

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Step 2.6.1

Add $2 x$ and $0$ .

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

Step 2.6.2

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

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

Step 2.6.3

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

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