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Calculus Examples
f(x)=2cos(x)+sin(2x)
Step 1
Step 1.1
By the Sum Rule, the derivative of 2cos(x)+sin(2x) with respect to x is ddx[2cos(x)]+ddx[sin(2x)].
ddx[2cos(x)]+ddx[sin(2x)]
Step 1.2
Evaluate ddx[2cos(x)].
Step 1.2.1
Since 2 is constant with respect to x, the derivative of 2cos(x) with respect to x is 2ddx[cos(x)].
2ddx[cos(x)]+ddx[sin(2x)]
Step 1.2.2
The derivative of cos(x) with respect to x is -sin(x).
2(-sin(x))+ddx[sin(2x)]
Step 1.2.3
Multiply -1 by 2.
-2sin(x)+ddx[sin(2x)]
-2sin(x)+ddx[sin(2x)]
Step 1.3
Evaluate ddx[sin(2x)].
Step 1.3.1
Differentiate using the chain rule, which states that ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=sin(x) and g(x)=2x.
Step 1.3.1.1
To apply the Chain Rule, set u as 2x.
-2sin(x)+ddu[sin(u)]ddx[2x]
Step 1.3.1.2
The derivative of sin(u) with respect to u is cos(u).
-2sin(x)+cos(u)ddx[2x]
Step 1.3.1.3
Replace all occurrences of u with 2x.
-2sin(x)+cos(2x)ddx[2x]
-2sin(x)+cos(2x)ddx[2x]
Step 1.3.2
Since 2 is constant with respect to x, the derivative of 2x with respect to x is 2ddx[x].
-2sin(x)+cos(2x)(2ddx[x])
Step 1.3.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
-2sin(x)+cos(2x)(2⋅1)
Step 1.3.4
Multiply 2 by 1.
-2sin(x)+cos(2x)⋅2
Step 1.3.5
Move 2 to the left of cos(2x).
-2sin(x)+2cos(2x)
-2sin(x)+2cos(2x)
-2sin(x)+2cos(2x)
Step 2
Step 2.1
By the Sum Rule, the derivative of -2sin(x)+2cos(2x) with respect to x is ddx[-2sin(x)]+ddx[2cos(2x)].
f′′(x)=ddx(-2sin(x))+ddx(2cos(2x))
Step 2.2
Evaluate ddx[-2sin(x)].
Step 2.2.1
Since -2 is constant with respect to x, the derivative of -2sin(x) with respect to x is -2ddx[sin(x)].
f′′(x)=-2ddxsin(x)+ddx(2cos(2x))
Step 2.2.2
The derivative of sin(x) with respect to x is cos(x).
f′′(x)=-2cos(x)+ddx(2cos(2x))
f′′(x)=-2cos(x)+ddx(2cos(2x))
Step 2.3
Evaluate ddx[2cos(2x)].
Step 2.3.1
Since 2 is constant with respect to x, the derivative of 2cos(2x) with respect to x is 2ddx[cos(2x)].
f′′(x)=-2cos(x)+2ddx(cos(2x))
Step 2.3.2
Differentiate using the chain rule, which states that ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=cos(x) and g(x)=2x.
Step 2.3.2.1
To apply the Chain Rule, set u as 2x.
f′′(x)=-2cos(x)+2(ddu(cos(u))ddx(2x))
Step 2.3.2.2
The derivative of cos(u) with respect to u is -sin(u).
f′′(x)=-2cos(x)+2(-sin(u)ddx(2x))
Step 2.3.2.3
Replace all occurrences of u with 2x.
f′′(x)=-2cos(x)+2(-sin(2x)ddx(2x))
f′′(x)=-2cos(x)+2(-sin(2x)ddx(2x))
Step 2.3.3
Since 2 is constant with respect to x, the derivative of 2x with respect to x is 2ddx[x].
f′′(x)=-2cos(x)+2(-sin(2x)(2ddx(x)))
Step 2.3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
f′′(x)=-2cos(x)+2(-sin(2x)(2⋅1))
Step 2.3.5
Multiply 2 by 1.
f′′(x)=-2cos(x)+2(-sin(2x)⋅2)
Step 2.3.6
Multiply 2 by -1.
f′′(x)=-2cos(x)+2(-2sin(2x))
Step 2.3.7
Multiply -2 by 2.
f′′(x)=-2cos(x)-4sin(2x)
f′′(x)=-2cos(x)-4sin(2x)
f′′(x)=-2cos(x)-4sin(2x)
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
-2sin(x)+2cos(2x)=0
Step 4
Step 4.1
Use the double-angle identity to transform cos(2x) to 1-2sin2(x).
-2sin(x)+2(1-2sin2(x))=0
Step 4.2
Apply the distributive property.
-2sin(x)+2⋅1+2(-2sin2(x))=0
Step 4.3
Multiply 2 by 1.
-2sin(x)+2+2(-2sin2(x))=0
Step 4.4
Multiply -2 by 2.
-2sin(x)+2-4sin2(x)=0
-2sin(x)+2-4sin2(x)=0
Step 5
Step 5.1
Factor 2 out of -2sin(x)+2-4sin2(x).
Step 5.1.1
Factor 2 out of -2sin(x).
2(-sin(x))+2-4sin2(x)=0
Step 5.1.2
Factor 2 out of 2.
2(-sin(x))+2(1)-4sin2(x)=0
Step 5.1.3
Factor 2 out of -4sin2(x).
2(-sin(x))+2(1)+2(-2sin2(x))=0
Step 5.1.4
Factor 2 out of 2(-sin(x))+2(1).
2(-sin(x)+1)+2(-2sin2(x))=0
Step 5.1.5
Factor 2 out of 2(-sin(x)+1)+2(-2sin2(x)).
2(-sin(x)+1-2sin2(x))=0
2(-sin(x)+1-2sin2(x))=0
Step 5.2
Factor.
Step 5.2.1
Factor by grouping.
Step 5.2.1.1
Reorder terms.
2(-2sin2(x)-sin(x)+1)=0
Step 5.2.1.2
For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=-2⋅1=-2 and whose sum is b=-1.
Step 5.2.1.2.1
Factor -1 out of -sin(x).
2(-2sin2(x)-sin(x)+1)=0
Step 5.2.1.2.2
Rewrite -1 as 1 plus -2
2(-2sin2(x)+(1-2)sin(x)+1)=0
Step 5.2.1.2.3
Apply the distributive property.
2(-2sin2(x)+1sin(x)-2sin(x)+1)=0
Step 5.2.1.2.4
Multiply sin(x) by 1.
2(-2sin2(x)+sin(x)-2sin(x)+1)=0
2(-2sin2(x)+sin(x)-2sin(x)+1)=0
Step 5.2.1.3
Factor out the greatest common factor from each group.
Step 5.2.1.3.1
Group the first two terms and the last two terms.
2(-2sin2(x)+sin(x)-2sin(x)+1)=0
Step 5.2.1.3.2
Factor out the greatest common factor (GCF) from each group.
2(sin(x)(-2sin(x)+1)+1(-2sin(x)+1))=0
2(sin(x)(-2sin(x)+1)+1(-2sin(x)+1))=0
Step 5.2.1.4
Factor the polynomial by factoring out the greatest common factor, -2sin(x)+1.
2((-2sin(x)+1)(sin(x)+1))=0
2((-2sin(x)+1)(sin(x)+1))=0
Step 5.2.2
Remove unnecessary parentheses.
2(-2sin(x)+1)(sin(x)+1)=0
2(-2sin(x)+1)(sin(x)+1)=0
2(-2sin(x)+1)(sin(x)+1)=0
Step 6
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
-2sin(x)+1=0
sin(x)+1=0
Step 7
Step 7.1
Set -2sin(x)+1 equal to 0.
-2sin(x)+1=0
Step 7.2
Solve -2sin(x)+1=0 for x.
Step 7.2.1
Subtract 1 from both sides of the equation.
-2sin(x)=-1
Step 7.2.2
Divide each term in -2sin(x)=-1 by -2 and simplify.
Step 7.2.2.1
Divide each term in -2sin(x)=-1 by -2.
-2sin(x)-2=-1-2
Step 7.2.2.2
Simplify the left side.
Step 7.2.2.2.1
Cancel the common factor of -2.
Step 7.2.2.2.1.1
Cancel the common factor.
-2sin(x)-2=-1-2
Step 7.2.2.2.1.2
Divide sin(x) by 1.
sin(x)=-1-2
sin(x)=-1-2
sin(x)=-1-2
Step 7.2.2.3
Simplify the right side.
Step 7.2.2.3.1
Dividing two negative values results in a positive value.
sin(x)=12
sin(x)=12
sin(x)=12
Step 7.2.3
Take the inverse sine of both sides of the equation to extract x from inside the sine.
x=arcsin(12)
Step 7.2.4
Simplify the right side.
Step 7.2.4.1
The exact value of arcsin(12) is π6.
x=π6
x=π6
Step 7.2.5
The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from π to find the solution in the second quadrant.
x=π-π6
Step 7.2.6
Simplify π-π6.
Step 7.2.6.1
To write π as a fraction with a common denominator, multiply by 66.
x=π⋅66-π6
Step 7.2.6.2
Combine fractions.
Step 7.2.6.2.1
Combine π and 66.
x=π⋅66-π6
Step 7.2.6.2.2
Combine the numerators over the common denominator.
x=π⋅6-π6
x=π⋅6-π6
Step 7.2.6.3
Simplify the numerator.
Step 7.2.6.3.1
Move 6 to the left of π.
x=6⋅π-π6
Step 7.2.6.3.2
Subtract π from 6π.
x=5π6
x=5π6
x=5π6
Step 7.2.7
The solution to the equation x=π6.
x=π6,5π6
x=π6,5π6
x=π6,5π6
Step 8
Step 8.1
Set sin(x)+1 equal to 0.
sin(x)+1=0
Step 8.2
Solve sin(x)+1=0 for x.
Step 8.2.1
Subtract 1 from both sides of the equation.
sin(x)=-1
Step 8.2.2
Take the inverse sine of both sides of the equation to extract x from inside the sine.
x=arcsin(-1)
Step 8.2.3
Simplify the right side.
Step 8.2.3.1
The exact value of arcsin(-1) is -π2.
x=-π2
x=-π2
Step 8.2.4
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from 2π, to find a reference angle. Next, add this reference angle to π to find the solution in the third quadrant.
x=2π+π2+π
Step 8.2.5
Simplify the expression to find the second solution.
Step 8.2.5.1
Subtract 2π from 2π+π2+π.
x=2π+π2+π-2π
Step 8.2.5.2
The resulting angle of 3π2 is positive, less than 2π, and coterminal with 2π+π2+π.
x=3π2
x=3π2
Step 8.2.6
The solution to the equation x=-π2.
x=-π2,3π2
x=-π2,3π2
x=-π2,3π2
Step 9
The final solution is all the values that make 2(-2sin(x)+1)(sin(x)+1)=0 true.
x=π6,5π6,-π2,3π2
Step 10
Evaluate the second derivative at x=π6. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
-2cos(π6)-4sin(2(π6))
Step 11
Step 11.1
Simplify each term.
Step 11.1.1
The exact value of cos(π6) is √32.
-2√32-4sin(2(π6))
Step 11.1.2
Cancel the common factor of 2.
Step 11.1.2.1
Factor 2 out of -2.
2(-1)√32-4sin(2(π6))
Step 11.1.2.2
Cancel the common factor.
2⋅-1√32-4sin(2(π6))
Step 11.1.2.3
Rewrite the expression.
-1√3-4sin(2(π6))
-1√3-4sin(2(π6))
Step 11.1.3
Rewrite -1√3 as -√3.
-√3-4sin(2(π6))
Step 11.1.4
Cancel the common factor of 2.
Step 11.1.4.1
Factor 2 out of 6.
-√3-4sin(2π2(3))
Step 11.1.4.2
Cancel the common factor.
-√3-4sin(2π2⋅3)
Step 11.1.4.3
Rewrite the expression.
-√3-4sin(π3)
-√3-4sin(π3)
Step 11.1.5
The exact value of sin(π3) is √32.
-√3-4√32
Step 11.1.6
Cancel the common factor of 2.
Step 11.1.6.1
Factor 2 out of -4.
-√3+2(-2)√32
Step 11.1.6.2
Cancel the common factor.
-√3+2⋅-2√32
Step 11.1.6.3
Rewrite the expression.
-√3-2√3
-√3-2√3
-√3-2√3
Step 11.2
Subtract 2√3 from -√3.
-3√3
-3√3
Step 12
x=π6 is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
x=π6 is a local maximum
Step 13
Step 13.1
Replace the variable x with π6 in the expression.
f(π6)=2cos(π6)+sin(2(π6))
Step 13.2
Simplify the result.
Step 13.2.1
Simplify each term.
Step 13.2.1.1
The exact value of cos(π6) is √32.
f(π6)=2(√32)+sin(2(π6))
Step 13.2.1.2
Cancel the common factor of 2.
Step 13.2.1.2.1
Cancel the common factor.
f(π6)=2(√32)+sin(2(π6))
Step 13.2.1.2.2
Rewrite the expression.
f(π6)=√3+sin(2(π6))
f(π6)=√3+sin(2(π6))
Step 13.2.1.3
Cancel the common factor of 2.
Step 13.2.1.3.1
Factor 2 out of 6.
f(π6)=√3+sin(2(π2(3)))
Step 13.2.1.3.2
Cancel the common factor.
f(π6)=√3+sin(2(π2⋅3))
Step 13.2.1.3.3
Rewrite the expression.
f(π6)=√3+sin(π3)
f(π6)=√3+sin(π3)
Step 13.2.1.4
The exact value of sin(π3) is √32.
f(π6)=√3+√32
f(π6)=√3+√32
Step 13.2.2
To write √3 as a fraction with a common denominator, multiply by 22.
f(π6)=√3⋅22+√32
Step 13.2.3
Combine fractions.
Step 13.2.3.1
Combine √3 and 22.
f(π6)=√3⋅22+√32
Step 13.2.3.2
Combine the numerators over the common denominator.
f(π6)=√3⋅2+√32
f(π6)=√3⋅2+√32
Step 13.2.4
Simplify the numerator.
Step 13.2.4.1
Move 2 to the left of √3.
f(π6)=2⋅√3+√32
Step 13.2.4.2
Add 2√3 and √3.
f(π6)=3√32
f(π6)=3√32
Step 13.2.5
The final answer is 3√32.
y=3√32
y=3√32
y=3√32
Step 14
Evaluate the second derivative at x=5π6. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
-2cos(5π6)-4sin(2(5π6))
Step 15
Step 15.1
Simplify each term.
Step 15.1.1
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 second quadrant.
-2(-cos(π6))-4sin(2(5π6))
Step 15.1.2
The exact value of cos(π6) is √32.
-2(-√32)-4sin(2(5π6))
Step 15.1.3
Cancel the common factor of 2.
Step 15.1.3.1
Move the leading negative in -√32 into the numerator.
-2-√32-4sin(2(5π6))
Step 15.1.3.2
Factor 2 out of -2.
2(-1)-√32-4sin(2(5π6))
Step 15.1.3.3
Cancel the common factor.
2⋅-1-√32-4sin(2(5π6))
Step 15.1.3.4
Rewrite the expression.
-1(-√3)-4sin(2(5π6))
-1(-√3)-4sin(2(5π6))
Step 15.1.4
Multiply -1 by -1.
1√3-4sin(2(5π6))
Step 15.1.5
Multiply √3 by 1.
√3-4sin(2(5π6))
Step 15.1.6
Cancel the common factor of 2.
Step 15.1.6.1
Factor 2 out of 6.
√3-4sin(25π2(3))
Step 15.1.6.2
Cancel the common factor.
√3-4sin(25π2⋅3)
Step 15.1.6.3
Rewrite the expression.
√3-4sin(5π3)
√3-4sin(5π3)
Step 15.1.7
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.
√3-4(-sin(π3))
Step 15.1.8
The exact value of sin(π3) is √32.
√3-4(-√32)
Step 15.1.9
Cancel the common factor of 2.
Step 15.1.9.1
Move the leading negative in -√32 into the numerator.
√3-4-√32
Step 15.1.9.2
Factor 2 out of -4.
√3+2(-2)-√32
Step 15.1.9.3
Cancel the common factor.
√3+2⋅-2-√32
Step 15.1.9.4
Rewrite the expression.
√3-2(-√3)
√3-2(-√3)
Step 15.1.10
Multiply -1 by -2.
√3+2√3
√3+2√3
Step 15.2
Add √3 and 2√3.
3√3
3√3
Step 16
x=5π6 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=5π6 is a local minimum
Step 17
Step 17.1
Replace the variable x with 5π6 in the expression.
f(5π6)=2cos(5π6)+sin(2(5π6))
Step 17.2
Simplify the result.
Step 17.2.1
Simplify each term.
Step 17.2.1.1
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 second quadrant.
f(5π6)=2(-cos(π6))+sin(2(5π6))
Step 17.2.1.2
The exact value of cos(π6) is √32.
f(5π6)=2(-√32)+sin(2(5π6))
Step 17.2.1.3
Cancel the common factor of 2.
Step 17.2.1.3.1
Move the leading negative in -√32 into the numerator.
f(5π6)=2(-√32)+sin(2(5π6))
Step 17.2.1.3.2
Cancel the common factor.
f(5π6)=2(-√32)+sin(2(5π6))
Step 17.2.1.3.3
Rewrite the expression.
f(5π6)=-√3+sin(2(5π6))
f(5π6)=-√3+sin(2(5π6))
Step 17.2.1.4
Cancel the common factor of 2.
Step 17.2.1.4.1
Factor 2 out of 6.
f(5π6)=-√3+sin(2(5π2(3)))
Step 17.2.1.4.2
Cancel the common factor.
f(5π6)=-√3+sin(2(5π2⋅3))
Step 17.2.1.4.3
Rewrite the expression.
f(5π6)=-√3+sin(5π3)
f(5π6)=-√3+sin(5π3)
Step 17.2.1.5
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.
f(5π6)=-√3-sin(π3)
Step 17.2.1.6
The exact value of sin(π3) is √32.
f(5π6)=-√3-√32
f(5π6)=-√3-√32
Step 17.2.2
To write -√3 as a fraction with a common denominator, multiply by 22.
f(5π6)=-√3⋅22-√32
Step 17.2.3
Combine fractions.
Step 17.2.3.1
Combine -√3 and 22.
f(5π6)=-√3⋅22-√32
Step 17.2.3.2
Combine the numerators over the common denominator.
f(5π6)=-√3⋅2-√32
f(5π6)=-√3⋅2-√32
Step 17.2.4
Simplify the numerator.
Step 17.2.4.1
Multiply 2 by -1.
f(5π6)=-2√3-√32
Step 17.2.4.2
Subtract √3 from -2√3.
f(5π6)=-3√32
f(5π6)=-3√32
Step 17.2.5
Move the negative in front of the fraction.
f(5π6)=-3√32
Step 17.2.6
The final answer is -3√32.
y=-3√32
y=-3√32
y=-3√32
Step 18
Evaluate the second derivative at x=-π2. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
-2cos(-π2)-4sin(2(-π2))
Step 19
Step 19.1
Simplify each term.
Step 19.1.1
Add full rotations of 2π until the angle is greater than or equal to 0 and less than 2π.
-2cos(3π2)-4sin(2(-π2))
Step 19.1.2
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.
-2cos(π2)-4sin(2(-π2))
Step 19.1.3
The exact value of cos(π2) is 0.
-2⋅0-4sin(2(-π2))
Step 19.1.4
Multiply -2 by 0.
0-4sin(2(-π2))
Step 19.1.5
Cancel the common factor of 2.
Step 19.1.5.1
Move the leading negative in -π2 into the numerator.
0-4sin(2-π2)
Step 19.1.5.2
Cancel the common factor.
0-4sin(2-π2)
Step 19.1.5.3
Rewrite the expression.
0-4sin(-π)
0-4sin(-π)
Step 19.1.6
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.
0-4sin(0)
Step 19.1.7
The exact value of sin(0) is 0.
0-4⋅0
Step 19.1.8
Multiply -4 by 0.
0+0
0+0
Step 19.2
Add 0 and 0.
0
0
Step 20
Step 20.1
Split (-∞,∞) into separate intervals around the x values that make the first derivative 0 or undefined.
(-∞,-π2)∪(-π2,π6)∪(π6,5π6)∪(5π6,3π2)∪(3π2,∞)
Step 20.2
Substitute any number, such as -4, from the interval (-∞,-π2) in the first derivative -2sin(x)+2cos(2x) to check if the result is negative or positive.
Step 20.2.1
Replace the variable x with -4 in the expression.
f′(-4)=-2sin(-4)+2cos(2(-4))
Step 20.2.2
Simplify the result.
Step 20.2.2.1
Simplify each term.
Step 20.2.2.1.1
Evaluate sin(-4).
f′(-4)=-2⋅0.75680249+2cos(2(-4))
Step 20.2.2.1.2
Multiply -2 by 0.75680249.
f′(-4)=-1.51360499+2cos(2(-4))
Step 20.2.2.1.3
Multiply 2 by -4.
f′(-4)=-1.51360499+2cos(-8)
Step 20.2.2.1.4
Evaluate cos(-8).
f′(-4)=-1.51360499+2⋅-0.14550003
Step 20.2.2.1.5
Multiply 2 by -0.14550003.
f′(-4)=-1.51360499-0.29100006
f′(-4)=-1.51360499-0.29100006
Step 20.2.2.2
Subtract 0.29100006 from -1.51360499.
f′(-4)=-1.80460505
Step 20.2.2.3
The final answer is -1.80460505.
-1.80460505
-1.80460505
-1.80460505
Step 20.3
Substitute any number, such as 0, from the interval (-π2,π6) in the first derivative -2sin(x)+2cos(2x) to check if the result is negative or positive.
Step 20.3.1
Replace the variable x with 0 in the expression.
f′(0)=-2sin(0)+2cos(2(0))
Step 20.3.2
Simplify the result.
Step 20.3.2.1
Simplify each term.
Step 20.3.2.1.1
The exact value of sin(0) is 0.
f′(0)=-2⋅0+2cos(2(0))
Step 20.3.2.1.2
Multiply -2 by 0.
f′(0)=0+2cos(2(0))
Step 20.3.2.1.3
Multiply 2 by 0.
f′(0)=0+2cos(0)
Step 20.3.2.1.4
The exact value of cos(0) is 1.
f′(0)=0+2⋅1
Step 20.3.2.1.5
Multiply 2 by 1.
f′(0)=0+2
f′(0)=0+2
Step 20.3.2.2
Add 0 and 2.
f′(0)=2
Step 20.3.2.3
The final answer is 2.
2
2
2
Step 20.4
Substitute any number, such as 2, from the interval (π6,5π6) in the first derivative -2sin(x)+2cos(2x) to check if the result is negative or positive.
Step 20.4.1
Replace the variable x with 2 in the expression.
f′(2)=-2sin(2)+2cos(2(2))
Step 20.4.2
Simplify the result.
Step 20.4.2.1
Simplify each term.
Step 20.4.2.1.1
Evaluate sin(2).
f′(2)=-2⋅0.90929742+2cos(2(2))
Step 20.4.2.1.2
Multiply -2 by 0.90929742.
f′(2)=-1.81859485+2cos(2(2))
Step 20.4.2.1.3
Multiply 2 by 2.
f′(2)=-1.81859485+2cos(4)
Step 20.4.2.1.4
Evaluate cos(4).
f′(2)=-1.81859485+2⋅-0.65364362
Step 20.4.2.1.5
Multiply 2 by -0.65364362.
f′(2)=-1.81859485-1.30728724
f′(2)=-1.81859485-1.30728724
Step 20.4.2.2
Subtract 1.30728724 from -1.81859485.
f′(2)=-3.12588209
Step 20.4.2.3
The final answer is -3.12588209.
-3.12588209
-3.12588209
-3.12588209
Step 20.5
Substitute any number, such as 4, from the interval (5π6,3π2) in the first derivative -2sin(x)+2cos(2x) to check if the result is negative or positive.
Step 20.5.1
Replace the variable x with 4 in the expression.
f′(4)=-2sin(4)+2cos(2(4))
Step 20.5.2
Simplify the result.
Step 20.5.2.1
Simplify each term.
Step 20.5.2.1.1
Evaluate sin(4).
f′(4)=-2⋅-0.75680249+2cos(2(4))
Step 20.5.2.1.2
Multiply -2 by -0.75680249.
f′(4)=1.51360499+2cos(2(4))
Step 20.5.2.1.3
Multiply 2 by 4.
f′(4)=1.51360499+2cos(8)
Step 20.5.2.1.4
Evaluate cos(8).
f′(4)=1.51360499+2⋅-0.14550003
Step 20.5.2.1.5
Multiply 2 by -0.14550003.
f′(4)=1.51360499-0.29100006
f′(4)=1.51360499-0.29100006
Step 20.5.2.2
Subtract 0.29100006 from 1.51360499.
f′(4)=1.22260492
Step 20.5.2.3
The final answer is 1.22260492.
1.22260492
1.22260492
1.22260492
Step 20.6
Substitute any number, such as 7, from the interval (3π2,∞) in the first derivative -2sin(x)+2cos(2x) to check if the result is negative or positive.
Step 20.6.1
Replace the variable x with 7 in the expression.
f′(7)=-2sin(7)+2cos(2(7))
Step 20.6.2
Simplify the result.
Step 20.6.2.1
Simplify each term.
Step 20.6.2.1.1
Evaluate sin(7).
f′(7)=-2⋅0.65698659+2cos(2(7))
Step 20.6.2.1.2
Multiply -2 by 0.65698659.
f′(7)=-1.31397319+2cos(2(7))
Step 20.6.2.1.3
Multiply 2 by 7.
f′(7)=-1.31397319+2cos(14)
Step 20.6.2.1.4
Evaluate cos(14).
f′(7)=-1.31397319+2⋅0.13673721
Step 20.6.2.1.5
Multiply 2 by 0.13673721.
f′(7)=-1.31397319+0.27347443
f′(7)=-1.31397319+0.27347443
Step 20.6.2.2
Add -1.31397319 and 0.27347443.
f′(7)=-1.04049876
Step 20.6.2.3
The final answer is -1.04049876.
-1.04049876
-1.04049876
-1.04049876
Step 20.7
Since the first derivative changed signs from negative to positive around x=-π2, then x=-π2 is a local minimum.
x=-π2 is a local minimum
Step 20.8
Since the first derivative changed signs from positive to negative around x=π6, then x=π6 is a local maximum.
x=π6 is a local maximum
Step 20.9
Since the first derivative changed signs from negative to positive around x=5π6, then x=5π6 is a local minimum.
x=5π6 is a local minimum
Step 20.10
Since the first derivative changed signs from positive to negative around x=3π2, then x=3π2 is a local maximum.
x=3π2 is a local maximum
Step 20.11
These are the local extrema for f(x)=2cos(x)+sin(2x).
x=-π2 is a local minimum
x=π6 is a local maximum
x=5π6 is a local minimum
x=3π2 is a local maximum
x=-π2 is a local minimum
x=π6 is a local maximum
x=5π6 is a local minimum
x=3π2 is a local maximum
Step 21