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Calculus Examples
Step 1
Step 1.1
Differentiate using the chain rule, which states that is where and .
Step 1.1.1
To apply the Chain Rule, set as .
Step 1.1.2
The derivative of with respect to is .
Step 1.1.3
Replace all occurrences of with .
Step 1.2
Differentiate.
Step 1.2.1
Since is constant with respect to , the derivative of with respect to is .
Step 1.2.2
Differentiate using the Power Rule which states that is where .
Step 1.2.3
Simplify the expression.
Step 1.2.3.1
Multiply by .
Step 1.2.3.2
Move to the left of .
Step 2
Step 2.1
Since is constant with respect to , the derivative of with respect to is .
Step 2.2
Differentiate using the chain rule, which states that is where and .
Step 2.2.1
To apply the Chain Rule, set as .
Step 2.2.2
The derivative of with respect to is .
Step 2.2.3
Replace all occurrences of with .
Step 2.3
Differentiate.
Step 2.3.1
Multiply by .
Step 2.3.2
Since is constant with respect to , the derivative of with respect to is .
Step 2.3.3
Multiply by .
Step 2.3.4
Differentiate using the Power Rule which states that is where .
Step 2.3.5
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
Divide each term in by .
Step 4.2
Simplify the left side.
Step 4.2.1
Cancel the common factor of .
Step 4.2.1.1
Cancel the common factor.
Step 4.2.1.2
Divide by .
Step 4.3
Simplify the right side.
Step 4.3.1
Divide by .
Step 5
Take the inverse cosine of both sides of the equation to extract from inside the cosine.
Step 6
Step 6.1
The exact value of is .
Step 7
Step 7.1
Divide each term in by .
Step 7.2
Simplify the left side.
Step 7.2.1
Cancel the common factor of .
Step 7.2.1.1
Cancel the common factor.
Step 7.2.1.2
Divide by .
Step 7.3
Simplify the right side.
Step 7.3.1
Multiply the numerator by the reciprocal of the denominator.
Step 7.3.2
Multiply .
Step 7.3.2.1
Multiply by .
Step 7.3.2.2
Multiply by .
Step 8
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 9
Step 9.1
Simplify.
Step 9.1.1
To write as a fraction with a common denominator, multiply by .
Step 9.1.2
Combine and .
Step 9.1.3
Combine the numerators over the common denominator.
Step 9.1.4
Multiply by .
Step 9.1.5
Subtract from .
Step 9.2
Divide each term in by and simplify.
Step 9.2.1
Divide each term in by .
Step 9.2.2
Simplify the left side.
Step 9.2.2.1
Cancel the common factor of .
Step 9.2.2.1.1
Cancel the common factor.
Step 9.2.2.1.2
Divide by .
Step 9.2.3
Simplify the right side.
Step 9.2.3.1
Multiply the numerator by the reciprocal of the denominator.
Step 9.2.3.2
Multiply .
Step 9.2.3.2.1
Multiply by .
Step 9.2.3.2.2
Multiply by .
Step 10
The solution to the equation .
Step 11
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 12
Step 12.1
Cancel the common factor of .
Step 12.1.1
Factor out of .
Step 12.1.2
Cancel the common factor.
Step 12.1.3
Rewrite the expression.
Step 12.2
The exact value of is .
Step 12.3
Multiply by .
Step 13
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
Step 14
Step 14.1
Replace the variable with in the expression.
Step 14.2
Simplify the result.
Step 14.2.1
Cancel the common factor of .
Step 14.2.1.1
Factor out of .
Step 14.2.1.2
Cancel the common factor.
Step 14.2.1.3
Rewrite the expression.
Step 14.2.2
The exact value of is .
Step 14.2.3
The final answer is .
Step 15
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 16
Step 16.1
Cancel the common factor of .
Step 16.1.1
Factor out of .
Step 16.1.2
Cancel the common factor.
Step 16.1.3
Rewrite the expression.
Step 16.2
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 16.3
The exact value of is .
Step 16.4
Multiply .
Step 16.4.1
Multiply by .
Step 16.4.2
Multiply by .
Step 17
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 18
Step 18.1
Replace the variable with in the expression.
Step 18.2
Simplify the result.
Step 18.2.1
Cancel the common factor of .
Step 18.2.1.1
Factor out of .
Step 18.2.1.2
Cancel the common factor.
Step 18.2.1.3
Rewrite the expression.
Step 18.2.2
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 18.2.3
The exact value of is .
Step 18.2.4
Multiply by .
Step 18.2.5
The final answer is .
Step 19
These are the local extrema for .
is a local maxima
is a local minima
Step 20