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Calculus Examples
y=-2cos(-x2)y=−2cos(−x2)
Step 1
Step 1.1
Since -2−2 is constant with respect to xx, the derivative of -2cos(-x2)−2cos(−x2) with respect to xx is -2ddx[cos(-x2)]−2ddx[cos(−x2)].
-2ddx[cos(-x2)]−2ddx[cos(−x2)]
Step 1.2
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=cos(x) and g(x)=-x2.
Step 1.2.1
To apply the Chain Rule, set u as -x2.
-2(ddu[cos(u)]ddx[-x2])
Step 1.2.2
The derivative of cos(u) with respect to u is -sin(u).
-2(-sin(u)ddx[-x2])
Step 1.2.3
Replace all occurrences of u with -x2.
-2(-sin(-x2)ddx[-x2])
-2(-sin(-x2)ddx[-x2])
Step 1.3
Differentiate.
Step 1.3.1
Multiply -1 by -2.
2(sin(-x2)ddx[-x2])
Step 1.3.2
Since -12 is constant with respect to x, the derivative of -x2 with respect to x is -12ddx[x].
2sin(-x2)(-12ddx[x])
Step 1.3.3
Simplify terms.
Step 1.3.3.1
Multiply -1 by 2.
-2sin(-x2)(12ddx[x])
Step 1.3.3.2
Combine 12 and -2.
-22sin(-x2)ddx[x]
Step 1.3.3.3
Combine -22 and sin(-x2).
-2sin(-x2)2ddx[x]
Step 1.3.3.4
Cancel the common factor of -2 and 2.
Step 1.3.3.4.1
Factor 2 out of -2sin(-x2).
2(-sin(-x2))2ddx[x]
Step 1.3.3.4.2
Cancel the common factors.
Step 1.3.3.4.2.1
Factor 2 out of 2.
2(-sin(-x2))2(1)ddx[x]
Step 1.3.3.4.2.2
Cancel the common factor.
2(-sin(-x2))2⋅1ddx[x]
Step 1.3.3.4.2.3
Rewrite the expression.
-sin(-x2)1ddx[x]
Step 1.3.3.4.2.4
Divide -sin(-x2) by 1.
-sin(-x2)ddx[x]
-sin(-x2)ddx[x]
-sin(-x2)ddx[x]
-sin(-x2)ddx[x]
Step 1.3.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
-sin(-x2)⋅1
Step 1.3.5
Multiply -1 by 1.
-sin(-x2)
-sin(-x2)
Step 1.4
Simplify.
Step 1.4.1
Since sin(-x2) is an odd function, rewrite sin(-x2) as -sin(x2).
--sin(x2)
Step 1.4.2
Multiply --sin(x2).
Step 1.4.2.1
Multiply -1 by -1.
1sin(x2)
Step 1.4.2.2
Multiply sin(x2) by 1.
sin(x2)
sin(x2)
sin(x2)
sin(x2)
Step 2
Step 2.1
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)=x2.
Step 2.1.1
To apply the Chain Rule, set u as x2.
f′′(x)=ddu(sin(u))ddx(x2)
Step 2.1.2
The derivative of sin(u) with respect to u is cos(u).
f′′(x)=cos(u)ddx(x2)
Step 2.1.3
Replace all occurrences of u with x2.
f′′(x)=cos(x2)ddx(x2)
f′′(x)=cos(x2)ddx(x2)
Step 2.2
Differentiate.
Step 2.2.1
Since 12 is constant with respect to x, the derivative of x2 with respect to x is 12ddx[x].
f′′(x)=cos(x2)(12⋅ddx(x))
Step 2.2.2
Combine 12 and cos(x2).
f′′(x)=cos(x2)2⋅ddx(x)
Step 2.2.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
f′′(x)=cos(x2)2⋅1
Step 2.2.4
Multiply cos(x2)2 by 1.
f′′(x)=cos(x2)2
f′′(x)=cos(x2)2
f′′(x)=cos(x2)2
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
sin(x2)=0
Step 4
Take the inverse sine of both sides of the equation to extract x from inside the sine.
x2=arcsin(0)
Step 5
Step 5.1
The exact value of arcsin(0) is 0.
x2=0
x2=0
Step 6
Set the numerator equal to zero.
x=0
Step 7
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.
x2=π-0
Step 8
Step 8.1
Multiply both sides of the equation by 2.
2x2=2(π-0)
Step 8.2
Simplify both sides of the equation.
Step 8.2.1
Simplify the left side.
Step 8.2.1.1
Cancel the common factor of 2.
Step 8.2.1.1.1
Cancel the common factor.
2x2=2(π-0)
Step 8.2.1.1.2
Rewrite the expression.
x=2(π-0)
x=2(π-0)
x=2(π-0)
Step 8.2.2
Simplify the right side.
Step 8.2.2.1
Subtract 0 from π.
x=2π
x=2π
x=2π
x=2π
Step 9
The solution to the equation x2=0.
x=0,2π
Step 10
Evaluate the second derivative at x=0. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
cos(02)2
Step 11
Step 11.1
Cancel the common factor of 0 and 2.
Step 11.1.1
Factor 2 out of 0.
cos(2(0)2)2
Step 11.1.2
Cancel the common factors.
Step 11.1.2.1
Factor 2 out of 2.
cos(2⋅02⋅1)2
Step 11.1.2.2
Cancel the common factor.
cos(2⋅02⋅1)2
Step 11.1.2.3
Rewrite the expression.
cos(01)2
Step 11.1.2.4
Divide 0 by 1.
cos(0)2
cos(0)2
cos(0)2
Step 11.2
The exact value of cos(0) is 1.
12
12
Step 12
x=0 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=0 is a local minimum
Step 13
Step 13.1
Replace the variable x with 0 in the expression.
f(0)=-2cos(-02)
Step 13.2
Simplify the result.
Step 13.2.1
Divide 0 by 2.
f(0)=-2cos(-0)
Step 13.2.2
Multiply -1 by 0.
f(0)=-2cos(0)
Step 13.2.3
The exact value of cos(0) is 1.
f(0)=-2⋅1
Step 13.2.4
Multiply -2 by 1.
f(0)=-2
Step 13.2.5
The final answer is -2.
y=-2
y=-2
y=-2
Step 14
Evaluate the second derivative at x=2π. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
cos(2π2)2
Step 15
Step 15.1
Cancel the common factor of 2.
Step 15.1.1
Cancel the common factor.
cos(2π2)2
Step 15.1.2
Divide π by 1.
cos(π)2
cos(π)2
Step 15.2
Simplify the numerator.
Step 15.2.1
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.
-cos(0)2
Step 15.2.2
The exact value of cos(0) is 1.
-1⋅12
Step 15.2.3
Multiply -1 by 1.
-12
-12
Step 15.3
Move the negative in front of the fraction.
-12
-12
Step 16
x=2π is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.
x=2π is a local maximum
Step 17
Step 17.1
Replace the variable x with 2π in the expression.
f(2π)=-2cos(-2π2)
Step 17.2
Simplify the result.
Step 17.2.1
Cancel the common factor of 2.
Step 17.2.1.1
Cancel the common factor.
f(2π)=-2cos(-2π2)
Step 17.2.1.2
Divide π by 1.
f(2π)=-2cos(-π)
f(2π)=-2cos(-π)
Step 17.2.2
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(2π)=-2(-cos(0))
Step 17.2.3
The exact value of cos(0) is 1.
f(2π)=-2(-1⋅1)
Step 17.2.4
Multiply -2(-1⋅1).
Step 17.2.4.1
Multiply -1 by 1.
f(2π)=-2⋅-1
Step 17.2.4.2
Multiply -2 by -1.
f(2π)=2
f(2π)=2
Step 17.2.5
The final answer is 2.
y=2
y=2
y=2
Step 18
These are the local extrema for f(x)=-2cos(-x2).
(0,-2) is a local minima
(2π,2) is a local maxima
Step 19