# Calculus Examples

Find the first derivative.

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 to get .

Find the second derivative.

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 to get .

The derivative of with respect to is .

Divide each term in by .

Simplify the left side of the equation by cancelling the common factors.

Reduce the expression by cancelling the common factors.

Factor out of .

Cancel the common factor.

Rewrite the expression.

Move the negative one from the denominator of .

Simplify the expression.

Multiply by to get .

Rewrite as .

Divide by to get .

Substitute in to find the value of .

Replace the variable with in the expression.

Simplify the result.

Remove parentheses around .

Raising to any positive power yields .

Multiply by to get .

The final answer is .

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

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

Replace the variable with in the expression.

Simplify the result.

Multiply by to get .

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.

Replace the variable with in the expression.

Simplify the result.

Multiply by to get .

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.

Multiply by to get .

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

From to , the second derivative changes from increasing to decreasing. There is a valid inflection point at .

Inflection point at .

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 point in this case is .