# Calculus Examples

Find the first 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 .

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 .

Add to both sides of the equation.

Divide each term by and simplify.

Divide each term in by .

Cancel the common factor of .

Cancel the common factor.

Divide by .

Cancel the common factor of and .

Factor out of .

Cancel the common factors.

Factor out of .

Cancel the common factor.

Rewrite the expression.

Substitute in to find the value of .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Apply the product rule to .

One to any power is one.

Raise to the power of .

Combine and .

Apply the product rule to .

One to any power is one.

Raise to the power of .

Combine and .

Move the negative in front of the fraction.

To write as a fraction with a common denominator, multiply by .

Write each expression with a common denominator of , by multiplying each by an appropriate factor of .

Multiply and .

Multiply by .

Combine the numerators over the common denominator.

Simplify the numerator.

Multiply by .

Subtract from .

Move the negative in front of the fraction.

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 .

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.

Multiply by .

Subtract from .

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

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 .