# Calculus Examples

Find the second derivative.

Find the first derivative.

By the Sum Rule, the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Evaluate .

Since is constant with respect to , the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Multiply by .

Find the second derivative.

By the Sum Rule, the derivative of with respect to is .

Evaluate .

Since is constant with respect to , the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Multiply by .

Evaluate .

Since is constant with respect to , the derivative of with respect to is .

Differentiate using the Power Rule which states that is where .

Multiply by .

The second derivative of with respect to is .

Set the second derivative equal to then solve the equation .

Factor out of .

Factor out of .

Factor out of .

Factor out of .

Divide each term by and simplify.

Divide each term in by .

Reduce the expression by cancelling the common factors.

Cancel the common factor.

Divide by .

Divide by .

Set equal to and solve for .

Set the factor equal to .

Add to both sides of the equation.

The solution is the result of and .

Find the points where the second derivative is .

Substitute in to find the value of .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

Raising to any positive power yields .

Remove parentheses.

Raising to any positive power yields .

Multiply by .

Add and .

The final answer is .

The point found by substituting in is . This point can be an inflection point.

Substitute in to find the value of .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

Raise to the power of .

Remove parentheses.

Raise to the power of .

Multiply by .

Subtract from .

The final answer is .

The point found by substituting in is . This point can be an inflection point.

Determine the points that could be inflection points.

Split into intervals around the points that could potentially be inflection points.

Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Multiply by .

Add and .

The final answer is .

At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .

Increasing on since

Increasing on since

Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.

Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Multiply by .

Add and .

The final answer is .

At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .

Increasing on since

Increasing on since

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

One to any power is one.

Multiply by .

Multiply by .

Subtract from .

The final answer is .

At , the second derivative is . Since this is negative, the second derivative is decreasing on the interval

Decreasing on since

Decreasing on since

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Multiply by .

Add and .

The final answer is .

At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .

Increasing on since

Increasing on since

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

One to any power is one.

Multiply by .

Multiply by .

Subtract from .

The final answer is .

At , the second derivative is . Since this is negative, the second derivative is decreasing on the interval

Decreasing on since

Decreasing on since

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

Raise to the power of .

Multiply by .

Multiply by .

Subtract from .

The final answer is .

Increasing on since

Increasing on since

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 .

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Create intervals around the inflection points and the undefined values.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Multiply by .

Add and .

The final answer is .

The graph is concave up on the interval because is positive.

Concave up on since is positive

Concave up on since is positive

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

One to any power is one.

Multiply by .

Multiply by .

Subtract from .

The final answer is .

The graph is concave down on the interval because is negative.

Concave down on since is negative

Concave down on since is negative

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses.

Raise to the power of .

Multiply by .

Multiply by .

Subtract from .

The final answer is .

The graph is concave up on the interval because is positive.

Concave up on since is positive

Concave up on since is positive

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on since is positive

Concave down on since is negative

Concave up on since is positive