Calculus Examples

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

Step 1

Write $y = \ln (x^{2} + 1)$ as a function.

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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

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Step 2.1.1.1

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

$f' (x) = \frac{d}{d u} (\ln (u)) \frac{d}{d x} (x^{2} + 1)$

Step 2.1.1.2

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

$f' (x) = \frac{1}{u} \cdot \frac{d}{d x} (x^{2} + 1)$

Step 2.1.1.3

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

$f' (x) = \frac{1}{x^{2} + 1} \cdot \frac{d}{d x} (x^{2} + 1)$

Step 2.1.2

Differentiate.

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Step 2.1.2.1

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

$f' (x) = \frac{1}{x^{2} + 1} \cdot (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (1))$

Step 2.1.2.2

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

$f' (x) = \frac{1}{x^{2} + 1} \cdot (2 x + \frac{d}{d x} (1))$

Step 2.1.2.3

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

$f' (x) = \frac{1}{x^{2} + 1} \cdot (2 x + 0)$

Step 2.1.2.4

Combine fractions.

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Step 2.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = \frac{1}{x^{2} + 1} \cdot (2 x)$

Step 2.1.2.4.2

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

$f' (x) = \frac{2}{x^{2} + 1} \cdot x$

Step 2.1.2.4.3

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

$f' (x) = \frac{2 x}{x^{2} + 1}$

Step 2.2

Find the second derivative.

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Step 2.2.1

Since $2$ is constant with respect to $x$ , the derivative of $\frac{2 x}{x^{2} + 1}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{x}{x^{2} + 1}]$ .

$f'' (x) = 2 \frac{d}{d x} (\frac{x}{x^{2} + 1})$

Step 2.2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x^{2} + 1$ .

$f'' (x) = 2 (\frac{(x^{2} + 1) \frac{d}{d x} (x) - x \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}})$

Step 2.2.3

Differentiate.

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Step 2.2.3.1

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

$f'' (x) = 2 (\frac{(x^{2} + 1) \cdot 1 - x \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}})$

Step 2.2.3.2

Multiply $x^{2} + 1$ by $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - x \frac{d}{d x} (x^{2} + 1)}{{(x^{2} + 1)}^{2}})$

Step 2.2.3.3

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

$f'' (x) = 2 (\frac{x^{2} + 1 - x (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (1))}{{(x^{2} + 1)}^{2}})$

Step 2.2.3.4

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

$f'' (x) = 2 (\frac{x^{2} + 1 - x (2 x + \frac{d}{d x} (1))}{{(x^{2} + 1)}^{2}})$

Step 2.2.3.5

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

$f'' (x) = 2 (\frac{x^{2} + 1 - x (2 x + 0)}{{(x^{2} + 1)}^{2}})$

Step 2.2.3.6

Simplify the expression.

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Step 2.2.3.6.1

Add $2 x$ and $0$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - x (2 x)}{{(x^{2} + 1)}^{2}})$

Step 2.2.3.6.2

Multiply $2$ by $- 1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 x \cdot x}{{(x^{2} + 1)}^{2}})$

Step 2.2.4

Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 (x \cdot x)}{{(x^{2} + 1)}^{2}})$

Step 2.2.5

Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 (x \cdot x)}{{(x^{2} + 1)}^{2}})$

Step 2.2.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 x^{1 + 1}}{{(x^{2} + 1)}^{2}})$

Step 2.2.7

Add $1$ and $1$ .

$f'' (x) = 2 (\frac{x^{2} + 1 - 2 x^{2}}{{(x^{2} + 1)}^{2}})$

Step 2.2.8

Subtract $2 x^{2}$ from $x^{2}$ .

$f'' (x) = 2 (\frac{- x^{2} + 1}{{(x^{2} + 1)}^{2}})$

Step 2.2.9

Combine $2$ and $\frac{- x^{2} + 1}{{(x^{2} + 1)}^{2}}$ .

$f'' (x) = \frac{2 (- x^{2} + 1)}{{(x^{2} + 1)}^{2}}$

Step 2.2.10

Simplify.

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Step 2.2.10.1

Apply the distributive property.

$f'' (x) = \frac{2 (- x^{2}) + 2 \cdot 1}{{(x^{2} + 1)}^{2}}$

Step 2.2.10.2

Simplify each term.

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Step 2.2.10.2.1

Multiply $- 1$ by $2$ .

$f'' (x) = \frac{- 2 x^{2} + 2 \cdot 1}{{(x^{2} + 1)}^{2}}$

Step 2.2.10.2.2

Multiply $2$ by $1$ .

$f'' (x) = \frac{- 2 x^{2} + 2}{{(x^{2} + 1)}^{2}}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{- 2 x^{2} + 2}{{(x^{2} + 1)}^{2}}$ .

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

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{- 2 x^{2} + 2}{{(x^{2} + 1)}^{2}} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

Set the numerator equal to zero.

$- 2 x^{2} + 2 = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

Subtract $2$ from both sides of the equation.

$- 2 x^{2} = - 2$

Step 3.3.2

Divide each term in $- 2 x^{2} = - 2$ by $- 2$ and simplify.

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Step 3.3.2.1

Divide each term in $- 2 x^{2} = - 2$ by $- 2$ .

$\frac{- 2 x^{2}}{- 2} = \frac{- 2}{- 2}$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $- 2$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{- 2 x^{2}}{- 2} = \frac{- 2}{- 2}$

Step 3.3.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 2}{- 2}$

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Divide $- 2$ by $- 2$ .

$x^{2} = 1$

Step 3.3.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Step 3.3.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 3.3.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.3.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 3.3.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 3.3.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

Substitute $1$ in $f (x) = \ln (x^{2} + 1)$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln ({(1)}^{2} + 1)$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

One to any power is one.

$f (1) = \ln (1 + 1)$

Step 4.1.2.2

Add $1$ and $1$ .

$f (1) = \ln (2)$

Step 4.1.2.3

The final answer is $\ln (2)$ .

$\ln (2)$

Step 4.2

The point found by substituting $1$ in $f (x) = \ln (x^{2} + 1)$ is $(1, \ln (2))$ . This point can be an inflection point.

$(1, \ln (2))$

Step 4.3

Substitute $- 1$ in $f (x) = \ln (x^{2} + 1)$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \ln ({(- 1)}^{2} + 1)$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Raise $- 1$ to the power of $2$ .

$f (- 1) = \ln (1 + 1)$

Step 4.3.2.2

Add $1$ and $1$ .

$f (- 1) = \ln (2)$

Step 4.3.2.3

The final answer is $\ln (2)$ .

$\ln (2)$

Step 4.4

The point found by substituting $- 1$ in $f (x) = \ln (x^{2} + 1)$ is $(- 1, \ln (2))$ . This point can be an inflection point.

$(- 1, \ln (2))$

Step 4.5

Determine the points that could be inflection points.

$(1, \ln (2)), (- 1, \ln (2))$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = \frac{- 2 {(- 1.1)}^{2} + 2}{{({(- 1.1)}^{2} + 1)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = \frac{- 2 \cdot 1.21 + 2}{{({(- 1.1)}^{2} + 1)}^{2}}$

Step 6.2.1.2

Multiply $- 2$ by $1.21$ .

$f'' (- 1.1) = \frac{- 2.42 + 2}{{({(- 1.1)}^{2} + 1)}^{2}}$

Step 6.2.1.3

Add $- 2.42$ and $2$ .

$f'' (- 1.1) = \frac{- 0.42}{{({(- 1.1)}^{2} + 1)}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = \frac{- 0.42}{{(1.21 + 1)}^{2}}$

Step 6.2.2.2

Add $1.21$ and $1$ .

$f'' (- 1.1) = \frac{- 0.42}{{2.21}^{2}}$

Step 6.2.2.3

Raise $2.21$ to the power of $2$ .

$f'' (- 1.1) = \frac{- 0.42}{4.8841}$

Step 6.2.3

Divide $- 0.42$ by $4.8841$ .

$f'' (- 1.1) = - 0.08599332$

Step 6.2.4

The final answer is $- 0.08599332$ .

$- 0.08599332$

Step 6.3

At $- 1.1$ , the second derivative is $- 0.08599332$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 1)$

Decreasing on $(- \infty, - 1)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(- 1, 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = \frac{- 2 {(0)}^{2} + 2}{{({(0)}^{2} + 1)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the numerator.

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Step 7.2.1.1

Raising $0$ to any positive power yields $0$ .

$f'' (0) = \frac{- 2 \cdot 0 + 2}{{(0^{2} + 1)}^{2}}$

Step 7.2.1.2

Multiply $- 2$ by $0$ .

$f'' (0) = \frac{0 + 2}{{(0^{2} + 1)}^{2}}$

Step 7.2.1.3

Add $0$ and $2$ .

$f'' (0) = \frac{2}{{(0^{2} + 1)}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

Raising $0$ to any positive power yields $0$ .

$f'' (0) = \frac{2}{{(0 + 1)}^{2}}$

Step 7.2.2.2

Add $0$ and $1$ .

$f'' (0) = \frac{2}{1^{2}}$

Step 7.2.2.3

One to any power is one.

$f'' (0) = \frac{2}{1}$

Step 7.2.3

Divide $2$ by $1$ .

$f'' (0) = 2$

Step 7.2.4

The final answer is $2$ .

$2$

Step 7.3

At $0$ , the second derivative is $2$ . Since this is positive, the second derivative is increasing on the interval $(- 1, 1)$ .

Increasing on $(- 1, 1)$ since $f'' (x) > 0$

Step 8

Substitute a value from the interval $(1, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $1.1$ in the expression.

$f'' (1.1) = \frac{- 2 {(1.1)}^{2} + 2}{{({(1.1)}^{2} + 1)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the numerator.

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Step 8.2.1.1

Raise $1.1$ to the power of $2$ .

$f'' (1.1) = \frac{- 2 \cdot 1.21 + 2}{{({1.1}^{2} + 1)}^{2}}$

Step 8.2.1.2

Multiply $- 2$ by $1.21$ .

$f'' (1.1) = \frac{- 2.42 + 2}{{({1.1}^{2} + 1)}^{2}}$

Step 8.2.1.3

Add $- 2.42$ and $2$ .

$f'' (1.1) = \frac{- 0.42}{{({1.1}^{2} + 1)}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

Raise $1.1$ to the power of $2$ .

$f'' (1.1) = \frac{- 0.42}{{(1.21 + 1)}^{2}}$

Step 8.2.2.2

Add $1.21$ and $1$ .

$f'' (1.1) = \frac{- 0.42}{{2.21}^{2}}$

Step 8.2.2.3

Raise $2.21$ to the power of $2$ .

$f'' (1.1) = \frac{- 0.42}{4.8841}$

Step 8.2.3

Divide $- 0.42$ by $4.8841$ .

$f'' (1.1) = - 0.08599332$

Step 8.2.4

The final answer is $- 0.08599332$ .

$- 0.08599332$

Step 8.3

At $1.1$ , the second derivative is $- 0.08599332$ . Since this is negative, the second derivative is decreasing on the interval $(1, \infty)$

Decreasing on $(1, \infty)$ since $f'' (x) < 0$

Step 9

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- 1, \ln (2)), (1, \ln (2))$ .

$(- 1, \ln (2)), (1, \ln (2))$

Step 10