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Calculus Examples
Step 1
Write as a function.
Step 2
Step 2.1
By the Sum Rule, the derivative of with respect to is .
Step 2.2
The derivative of with respect to is .
Step 2.3
The derivative of with respect to is .
Step 3
Step 3.1
By the Sum Rule, the derivative of with respect to is .
Step 3.2
The derivative of with respect to is .
Step 3.3
Evaluate .
Step 3.3.1
Since is constant with respect to , the derivative of with respect to is .
Step 3.3.2
The derivative of with respect to is .
Step 4
To find the local maximum and minimum values of the function, set the derivative equal to and solve.
Step 5
Divide each term in the equation by .
Step 6
Step 6.1
Cancel the common factor.
Step 6.2
Rewrite the expression.
Step 7
Separate fractions.
Step 8
Convert from to .
Step 9
Divide by .
Step 10
Separate fractions.
Step 11
Convert from to .
Step 12
Divide by .
Step 13
Multiply by .
Step 14
Subtract from both sides of the equation.
Step 15
Step 15.1
Divide each term in by .
Step 15.2
Simplify the left side.
Step 15.2.1
Dividing two negative values results in a positive value.
Step 15.2.2
Divide by .
Step 15.3
Simplify the right side.
Step 15.3.1
Divide by .
Step 16
Take the inverse tangent of both sides of the equation to extract from inside the tangent.
Step 17
Step 17.1
The exact value of is .
Step 18
The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from to find the solution in the fourth quadrant.
Step 19
Step 19.1
To write as a fraction with a common denominator, multiply by .
Step 19.2
Combine fractions.
Step 19.2.1
Combine and .
Step 19.2.2
Combine the numerators over the common denominator.
Step 19.3
Simplify the numerator.
Step 19.3.1
Move to the left of .
Step 19.3.2
Add and .
Step 20
The solution to the equation .
Step 21
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 22
Step 22.1
Simplify each term.
Step 22.1.1
The exact value of is .
Step 22.1.2
The exact value of is .
Step 22.2
Simplify terms.
Step 22.2.1
Combine the numerators over the common denominator.
Step 22.2.2
Subtract from .
Step 22.2.3
Cancel the common factor of and .
Step 22.2.3.1
Factor out of .
Step 22.2.3.2
Cancel the common factors.
Step 22.2.3.2.1
Factor out of .
Step 22.2.3.2.2
Cancel the common factor.
Step 22.2.3.2.3
Rewrite the expression.
Step 22.2.3.2.4
Divide by .
Step 23
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 24
Step 24.1
Replace the variable with in the expression.
Step 24.2
Simplify the result.
Step 24.2.1
Simplify each term.
Step 24.2.1.1
The exact value of is .
Step 24.2.1.2
The exact value of is .
Step 24.2.2
Simplify terms.
Step 24.2.2.1
Combine the numerators over the common denominator.
Step 24.2.2.2
Add and .
Step 24.2.2.3
Cancel the common factor of .
Step 24.2.2.3.1
Cancel the common factor.
Step 24.2.2.3.2
Divide by .
Step 24.2.3
The final answer is .
Step 25
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 26
Step 26.1
Simplify each term.
Step 26.1.1
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 third quadrant.
Step 26.1.2
The exact value of is .
Step 26.1.3
Multiply .
Step 26.1.3.1
Multiply by .
Step 26.1.3.2
Multiply by .
Step 26.1.4
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.
Step 26.1.5
The exact value of is .
Step 26.1.6
Multiply .
Step 26.1.6.1
Multiply by .
Step 26.1.6.2
Multiply by .
Step 26.2
Simplify terms.
Step 26.2.1
Combine the numerators over the common denominator.
Step 26.2.2
Add and .
Step 26.2.3
Cancel the common factor of .
Step 26.2.3.1
Cancel the common factor.
Step 26.2.3.2
Divide by .
Step 27
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 28
Step 28.1
Replace the variable with in the expression.
Step 28.2
Simplify the result.
Step 28.2.1
Simplify each term.
Step 28.2.1.1
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 third quadrant.
Step 28.2.1.2
The exact value of is .
Step 28.2.1.3
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.
Step 28.2.1.4
The exact value of is .
Step 28.2.2
Simplify terms.
Step 28.2.2.1
Combine the numerators over the common denominator.
Step 28.2.2.2
Subtract from .
Step 28.2.2.3
Cancel the common factor of and .
Step 28.2.2.3.1
Factor out of .
Step 28.2.2.3.2
Cancel the common factors.
Step 28.2.2.3.2.1
Factor out of .
Step 28.2.2.3.2.2
Cancel the common factor.
Step 28.2.2.3.2.3
Rewrite the expression.
Step 28.2.2.3.2.4
Divide by .
Step 28.2.3
The final answer is .
Step 29
These are the local extrema for .
is a local maxima
is a local minima
Step 30