Enter a problem...
Calculus Examples
Step 1
Step 1.1
By the Sum Rule, the derivative of with respect to is .
Step 1.2
Evaluate .
Step 1.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 1.2.2
Differentiate using the chain rule, which states that is where and .
Step 1.2.2.1
To apply the Chain Rule, set as .
Step 1.2.2.2
Differentiate using the Power Rule which states that is where .
Step 1.2.2.3
Replace all occurrences of with .
Step 1.2.3
The derivative of with respect to is .
Step 1.2.4
Multiply by .
Step 1.2.5
Multiply by .
Step 1.3
Evaluate .
Step 1.3.1
Since is constant with respect to , the derivative of with respect to is .
Step 1.3.2
The derivative of with respect to is .
Step 2
Step 2.1
By the Sum Rule, the derivative of with respect to is .
Step 2.2
Evaluate .
Step 2.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 2.2.2
Differentiate using the Product Rule which states that is where and .
Step 2.2.3
The derivative of with respect to is .
Step 2.2.4
The derivative of with respect to is .
Step 2.2.5
Raise to the power of .
Step 2.2.6
Raise to the power of .
Step 2.2.7
Use the power rule to combine exponents.
Step 2.2.8
Add and .
Step 2.2.9
Raise to the power of .
Step 2.2.10
Raise to the power of .
Step 2.2.11
Use the power rule to combine exponents.
Step 2.2.12
Add and .
Step 2.3
Evaluate .
Step 2.3.1
Since is constant with respect to , the derivative of with respect to is .
Step 2.3.2
The derivative of with respect to is .
Step 2.3.3
Multiply by .
Step 2.4
Simplify.
Step 2.4.1
Apply the distributive property.
Step 2.4.2
Multiply by .
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to and solve.
Step 4
Step 4.1
Factor out of .
Step 4.1.1
Factor out of .
Step 4.1.2
Factor out of .
Step 4.1.3
Factor out of .
Step 4.2
Rewrite as .
Step 5
If any individual factor on the left side of the equation is equal to , the entire expression will be equal to .
Step 6
Step 6.1
Set equal to .
Step 6.2
Solve for .
Step 6.2.1
Take the inverse cosine of both sides of the equation to extract from inside the cosine.
Step 6.2.2
Simplify the right side.
Step 6.2.2.1
The exact value of is .
Step 6.2.3
The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from to find the solution in the fourth quadrant.
Step 6.2.4
Simplify .
Step 6.2.4.1
To write as a fraction with a common denominator, multiply by .
Step 6.2.4.2
Combine fractions.
Step 6.2.4.2.1
Combine and .
Step 6.2.4.2.2
Combine the numerators over the common denominator.
Step 6.2.4.3
Simplify the numerator.
Step 6.2.4.3.1
Multiply by .
Step 6.2.4.3.2
Subtract from .
Step 6.2.5
The solution to the equation .
Step 7
Step 7.1
Set equal to .
Step 7.2
Solve for .
Step 7.2.1
Add to both sides of the equation.
Step 7.2.2
Divide each term in by and simplify.
Step 7.2.2.1
Divide each term in by .
Step 7.2.2.2
Simplify the left side.
Step 7.2.2.2.1
Dividing two negative values results in a positive value.
Step 7.2.2.2.2
Divide by .
Step 7.2.2.3
Simplify the right side.
Step 7.2.2.3.1
Divide by .
Step 7.2.3
Take the inverse sine of both sides of the equation to extract from inside the sine.
Step 7.2.4
Simplify the right side.
Step 7.2.4.1
The exact value of is .
Step 7.2.5
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from , to find a reference angle. Next, add this reference angle to to find the solution in the third quadrant.
Step 7.2.6
Simplify the expression to find the second solution.
Step 7.2.6.1
Subtract from .
Step 7.2.6.2
The resulting angle of is positive, less than , and coterminal with .
Step 7.2.7
The solution to the equation .
Step 8
The final solution is all the values that make true.
Step 9
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.
Step 10
Step 10.1
Simplify each term.
Step 10.1.1
The exact value of is .
Step 10.1.2
Raising to any positive power yields .
Step 10.1.3
Multiply by .
Step 10.1.4
The exact value of is .
Step 10.1.5
One to any power is one.
Step 10.1.6
Multiply by .
Step 10.1.7
The exact value of is .
Step 10.1.8
Multiply by .
Step 10.2
Simplify by adding numbers.
Step 10.2.1
Add and .
Step 10.2.2
Add and .
Step 11
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
Step 12
Step 12.1
Replace the variable with in the expression.
Step 12.2
Simplify the result.
Step 12.2.1
Simplify each term.
Step 12.2.1.1
The exact value of is .
Step 12.2.1.2
Raising to any positive power yields .
Step 12.2.1.3
Multiply by .
Step 12.2.1.4
The exact value of is .
Step 12.2.1.5
Multiply by .
Step 12.2.2
Subtract from .
Step 12.2.3
The final answer is .
Step 13
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.
Step 14
Step 14.1
Simplify each term.
Step 14.1.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.
Step 14.1.2
The exact value of is .
Step 14.1.3
Raising to any positive power yields .
Step 14.1.4
Multiply by .
Step 14.1.5
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.
Step 14.1.6
The exact value of is .
Step 14.1.7
Multiply by .
Step 14.1.8
Raise to the power of .
Step 14.1.9
Multiply by .
Step 14.1.10
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.
Step 14.1.11
The exact value of is .
Step 14.1.12
Multiply .
Step 14.1.12.1
Multiply by .
Step 14.1.12.2
Multiply by .
Step 14.2
Simplify by adding and subtracting.
Step 14.2.1
Add and .
Step 14.2.2
Subtract from .
Step 15
Step 15.1
Split into separate intervals around the values that make the first derivative or undefined.
Step 15.2
Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.
Step 15.2.1
Replace the variable with in the expression.
Step 15.2.2
Simplify the result.
Step 15.2.2.1
Simplify each term.
Step 15.2.2.1.1
Evaluate .
Step 15.2.2.1.2
Multiply by .
Step 15.2.2.1.3
Evaluate .
Step 15.2.2.1.4
Multiply by .
Step 15.2.2.1.5
Evaluate .
Step 15.2.2.1.6
Multiply by .
Step 15.2.2.2
Add and .
Step 15.2.2.3
The final answer is .
Step 15.3
Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.
Step 15.3.1
Replace the variable with in the expression.
Step 15.3.2
Simplify the result.
Step 15.3.2.1
Simplify each term.
Step 15.3.2.1.1
The exact value of is .
Step 15.3.2.1.2
Multiply by .
Step 15.3.2.1.3
The exact value of is .
Step 15.3.2.1.4
Multiply by .
Step 15.3.2.1.5
The exact value of is .
Step 15.3.2.1.6
Multiply by .
Step 15.3.2.2
Subtract from .
Step 15.3.2.3
The final answer is .
Step 15.4
Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.
Step 15.4.1
Replace the variable with in the expression.
Step 15.4.2
Simplify the result.
Step 15.4.2.1
Simplify each term.
Step 15.4.2.1.1
Evaluate .
Step 15.4.2.1.2
Multiply by .
Step 15.4.2.1.3
Evaluate .
Step 15.4.2.1.4
Multiply by .
Step 15.4.2.1.5
Evaluate .
Step 15.4.2.1.6
Multiply by .
Step 15.4.2.2
Add and .
Step 15.4.2.3
The final answer is .
Step 15.5
Substitute any number, such as , from the interval in the first derivative to check if the result is negative or positive.
Step 15.5.1
Replace the variable with in the expression.
Step 15.5.2
Simplify the result.
Step 15.5.2.1
Simplify each term.
Step 15.5.2.1.1
Evaluate .
Step 15.5.2.1.2
Multiply by .
Step 15.5.2.1.3
Evaluate .
Step 15.5.2.1.4
Multiply by .
Step 15.5.2.1.5
Evaluate .
Step 15.5.2.1.6
Multiply by .
Step 15.5.2.2
Subtract from .
Step 15.5.2.3
The final answer is .
Step 15.6
Since the first derivative changed signs from positive to negative around , then is a local maximum.
is a local maximum
Step 15.7
Since the first derivative changed signs from negative to positive around , then is a local minimum.
is a local minimum
Step 15.8
Since the first derivative changed signs from positive to negative around , then is a local maximum.
is a local maximum
Step 15.9
These are the local extrema for .
is a local maximum
is a local minimum
is a local maximum
is a local maximum
is a local minimum
is a local maximum
Step 16