# Calculus Examples

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 .

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 in by .

Reduce the expression by cancelling the common factors.

Cancel the common factor.

Divide by .

Divide by .

Set the factor equal to .

Add to both sides of the equation.

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

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

First, use the positive value of the to find the first solution.

Next, use the negative value of the to find the second solution.

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

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.

Remove parentheses around .

Raising to any positive power yields .

Multiply by .

Subtract from .

is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

is a local maximum

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.

Remove parentheses around .

Rewrite as .

Multiply by .

Subtract from .

is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

is a local minimum

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.

Apply the product rule to .

Raise to the power of .

Multiply by .

Rewrite as .

Multiply by .

Subtract from .

is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

is a local minimum

These are the local extrema for .

is a local maximum

is a local minimum

is a local minimum