# Calculus Examples

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 .

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 .

To find the local maximum and minimum values of the function, set the derivative equal to and solve.

Factor out of .

Factor out of .

Factor out of .

Divide each term by and simplify.

Divide each term in by .

Cancel the common factor of .

Cancel the common factor.

Divide by .

Divide by .

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

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

Simplify the right side of the equation.

Rewrite as .

Pull terms out from under the radical, assuming positive real numbers.

is equal to .

Set the factor 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 .

The solution is the result of and .

Evaluate the second derivative at . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

Simplify each term.

Raising to any positive power yields .

Multiply by .

Multiply by .

Add and .

Split into separate intervals around the values that make the first derivative or undefined.

Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Raise to the power of .

Multiply by .

Subtract from .

The final answer is .

Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Raise to the power of .

Multiply by .

Subtract from .

The final answer is .

Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.

Replace the variable with in the expression.

Simplify the result.

Simplify each term.

Raise to the power of .

Multiply by .

Raise to the power of .

Multiply by .

Subtract from .

The final answer is .

Since the first derivative did not change signs around , this is not a local maximum or minimum.

Not a local maximum or minimum

Since the first derivative changed signs from negative to positive around , then is a local minimum.

is a local minimum

is a local minimum