# Calculus Examples

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 to get .

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 to get .

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

Add and to get .

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 to get .

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 to get .

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

Add and to get .

The derivative of with respect to is .

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 to get .

Divide by to get .

Set equal to and solve for .

Set the factor equal to .

Since does not contain the variable to solve for, move it to the right side of the equation by subtracting from both sides.

The solution is the result of and .

Substitute in to find the value of .

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses around .

Raising to any positive power yields .

Remove parentheses around .

Raising to any positive power yields .

Multiply by to get .

Multiply by to get .

Simplify by adding zeros.

Add and to get .

Add and to get .

Add and to get .

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.

Raise to the power of to get .

Raise to the power of to get .

Multiply by to get .

Multiply by to get .

Simplify by adding and subtracting.

Subtract from to get .

Add and to get .

Add and to get .

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.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of to get .

Multiply by to get .

Multiply by to get .

Subtract from 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.

Simplify each term.

Raise to the power of to get .

Multiply by to get .

Multiply by to get .

Subtract from 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.

Simplify each term.

Apply the product rule to .

Raise to the power of to get .

Multiply by to get .

Apply the product rule to .

One to any power is one.

Raise to the power of to get .

Cancel the common factor of .

Write as a fraction with denominator .

Factor out the greatest common factor .

Cancel the common factor.

Rewrite the expression.

Simplify.

Multiply and to get .

Divide by to get .

Cancel the common factor of .

Rewrite.

Write as a fraction with denominator .

Factor out the greatest common factor .

Cancel the common factor.

Rewrite the expression.

Reduce the expression by cancelling the common factors.

Cancel the common factor.

Rewrite the expression.

Simplify.

Multiply and to get .

Multiply by to get .

Divide by to get .

Subtract from 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

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 to get .

Multiply by to get .

Multiply by to get .

Subtract from 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.

Simplify each term.

Apply the product rule to .

Raise to the power of to get .

Multiply by to get .

Apply the product rule to .

One to any power is one.

Raise to the power of to get .

Cancel the common factor of .

Write as a fraction with denominator .

Factor out the greatest common factor .

Cancel the common factor.

Rewrite the expression.

Simplify.

Multiply and to get .

Divide by to get .

Cancel the common factor of .

Rewrite.

Write as a fraction with denominator .

Factor out the greatest common factor .

Cancel the common factor.

Rewrite the expression.

Reduce the expression by cancelling the common factors.

Cancel the common factor.

Rewrite the expression.

Simplify.

Multiply and to get .

Multiply by to get .

Divide by to get .

Subtract from 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

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Remove parentheses around .

Raise to the power of to get .

Multiply by to get .

Multiply by to get .

Add and to get .

The final answer is .

Increasing on since

Increasing on since

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

Inflection point at .

From to , the second derivative changes from decreasing to increasing. 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 points in this case are .