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Calculus Examples
f(x)=x4ln(x)f(x)=x4ln(x)
Step 1
Step 1.1
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x4 and g(x)=ln(x).
x4ddx[ln(x)]+ln(x)ddx[x4]
Step 1.2
The derivative of ln(x) with respect to x is 1x.
x41x+ln(x)ddx[x4]
Step 1.3
Differentiate using the Power Rule.
Step 1.3.1
Combine x4 and 1x.
x4x+ln(x)ddx[x4]
Step 1.3.2
Cancel the common factor of x4 and x.
Step 1.3.2.1
Factor x out of x4.
x⋅x3x+ln(x)ddx[x4]
Step 1.3.2.2
Cancel the common factors.
Step 1.3.2.2.1
Raise x to the power of 1.
x⋅x3x1+ln(x)ddx[x4]
Step 1.3.2.2.2
Factor x out of x1.
x⋅x3x⋅1+ln(x)ddx[x4]
Step 1.3.2.2.3
Cancel the common factor.
x⋅x3x⋅1+ln(x)ddx[x4]
Step 1.3.2.2.4
Rewrite the expression.
x31+ln(x)ddx[x4]
Step 1.3.2.2.5
Divide x3 by 1.
x3+ln(x)ddx[x4]
x3+ln(x)ddx[x4]
x3+ln(x)ddx[x4]
Step 1.3.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=4.
x3+ln(x)(4x3)
Step 1.3.4
Reorder terms.
x3+4x3ln(x)
x3+4x3ln(x)
x3+4x3ln(x)
Step 2
Step 2.1
Differentiate.
Step 2.1.1
By the Sum Rule, the derivative of x3+4x3ln(x) with respect to x is ddx[x3]+ddx[4x3ln(x)].
f′′(x)=ddx(x3)+ddx(4x3ln(x))
Step 2.1.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
f′′(x)=3x2+ddx(4x3ln(x))
f′′(x)=3x2+ddx(4x3ln(x))
Step 2.2
Evaluate ddx[4x3ln(x)].
Step 2.2.1
Since 4 is constant with respect to x, the derivative of 4x3ln(x) with respect to x is 4ddx[x3ln(x)].
f′′(x)=3x2+4ddx(x3ln(x))
Step 2.2.2
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x3 and g(x)=ln(x).
f′′(x)=3x2+4(x3ddx(ln(x))+ln(x)ddx(x3))
Step 2.2.3
The derivative of ln(x) with respect to x is 1x.
f′′(x)=3x2+4(x3(1x)+ln(x)ddx(x3))
Step 2.2.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
f′′(x)=3x2+4(x3(1x)+ln(x)(3x2))
Step 2.2.5
Combine x3 and 1x.
f′′(x)=3x2+4(x3x+ln(x)(3x2))
Step 2.2.6
Cancel the common factor of x3 and x.
Step 2.2.6.1
Factor x out of x3.
f′′(x)=3x2+4(x⋅x2x+ln(x)(3x2))
Step 2.2.6.2
Cancel the common factors.
Step 2.2.6.2.1
Raise x to the power of 1.
f′′(x)=3x2+4(x⋅x2x+ln(x)(3x2))
Step 2.2.6.2.2
Factor x out of x1.
f′′(x)=3x2+4(x⋅x2x⋅1+ln(x)(3x2))
Step 2.2.6.2.3
Cancel the common factor.
f′′(x)=3x2+4(x⋅x2x⋅1+ln(x)(3x2))
Step 2.2.6.2.4
Rewrite the expression.
f′′(x)=3x2+4(x21+ln(x)(3x2))
Step 2.2.6.2.5
Divide x2 by 1.
f′′(x)=3x2+4(x2+ln(x)(3x2))
f′′(x)=3x2+4(x2+ln(x)(3x2))
f′′(x)=3x2+4(x2+ln(x)(3x2))
f′′(x)=3x2+4(x2+ln(x)(3x2))
Step 2.3
Simplify.
Step 2.3.1
Apply the distributive property.
f′′(x)=3x2+4x2+4(ln(x)(3x2))
Step 2.3.2
Combine terms.
Step 2.3.2.1
Multiply 3 by 4.
f′′(x)=3x2+4x2+12(ln(x)(x2))
Step 2.3.2.2
Add 3x2 and 4x2.
f′′(x)=7x2+12ln(x)x2
f′′(x)=7x2+12ln(x)x2
Step 2.3.3
Reorder terms.
f′′(x)=7x2+12x2ln(x)
f′′(x)=7x2+12x2ln(x)
f′′(x)=7x2+12x2ln(x)
Step 3
To find the local maximum and minimum values of the function, set the derivative equal to 0 and solve.
x3+4x3ln(x)=0
Step 4
Step 4.1
Find the first derivative.
Step 4.1.1
Differentiate using the Product Rule which states that ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=x4 and g(x)=ln(x).
x4ddx[ln(x)]+ln(x)ddx[x4]
Step 4.1.2
The derivative of ln(x) with respect to x is 1x.
x41x+ln(x)ddx[x4]
Step 4.1.3
Differentiate using the Power Rule.
Step 4.1.3.1
Combine x4 and 1x.
x4x+ln(x)ddx[x4]
Step 4.1.3.2
Cancel the common factor of x4 and x.
Step 4.1.3.2.1
Factor x out of x4.
x⋅x3x+ln(x)ddx[x4]
Step 4.1.3.2.2
Cancel the common factors.
Step 4.1.3.2.2.1
Raise x to the power of 1.
x⋅x3x1+ln(x)ddx[x4]
Step 4.1.3.2.2.2
Factor x out of x1.
x⋅x3x⋅1+ln(x)ddx[x4]
Step 4.1.3.2.2.3
Cancel the common factor.
x⋅x3x⋅1+ln(x)ddx[x4]
Step 4.1.3.2.2.4
Rewrite the expression.
x31+ln(x)ddx[x4]
Step 4.1.3.2.2.5
Divide x3 by 1.
x3+ln(x)ddx[x4]
x3+ln(x)ddx[x4]
x3+ln(x)ddx[x4]
Step 4.1.3.3
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=4.
x3+ln(x)(4x3)
Step 4.1.3.4
Reorder terms.
f′(x)=x3+4x3ln(x)
f′(x)=x3+4x3ln(x)
f′(x)=x3+4x3ln(x)
Step 4.2
The first derivative of f(x) with respect to x is x3+4x3ln(x).
x3+4x3ln(x)
x3+4x3ln(x)
Step 5
Step 5.1
Set the first derivative equal to 0.
x3+4x3ln(x)=0
Step 5.2
Subtract x3 from both sides of the equation.
4x3ln(x)=-x3
Step 5.3
Divide each term in 4x3ln(x)=-x3 by 4x3 and simplify.
Step 5.3.1
Divide each term in 4x3ln(x)=-x3 by 4x3.
4x3ln(x)4x3=-x34x3
Step 5.3.2
Simplify the left side.
Step 5.3.2.1
Cancel the common factor of 4.
Step 5.3.2.1.1
Cancel the common factor.
4x3ln(x)4x3=-x34x3
Step 5.3.2.1.2
Rewrite the expression.
x3ln(x)x3=-x34x3
x3ln(x)x3=-x34x3
Step 5.3.2.2
Cancel the common factor of x3.
Step 5.3.2.2.1
Cancel the common factor.
x3ln(x)x3=-x34x3
Step 5.3.2.2.2
Divide ln(x) by 1.
ln(x)=-x34x3
ln(x)=-x34x3
ln(x)=-x34x3
Step 5.3.3
Simplify the right side.
Step 5.3.3.1
Cancel the common factor of x3.
Step 5.3.3.1.1
Cancel the common factor.
ln(x)=-x34x3
Step 5.3.3.1.2
Rewrite the expression.
ln(x)=-14
ln(x)=-14
Step 5.3.3.2
Move the negative in front of the fraction.
ln(x)=-14
ln(x)=-14
ln(x)=-14
Step 5.4
To solve for x, rewrite the equation using properties of logarithms.
eln(x)=e-14
Step 5.5
Rewrite ln(x)=-14 in exponential form using the definition of a logarithm. If x and b are positive real numbers and b≠1, then logb(x)=y is equivalent to by=x.
e-14=x
Step 5.6
Solve for x.
Step 5.6.1
Rewrite the equation as x=e-14.
x=e-14
Step 5.6.2
Rewrite the expression using the negative exponent rule b-n=1bn.
x=1e14
x=1e14
x=1e14
Step 6
Step 6.1
Set the argument in ln(x) less than or equal to 0 to find where the expression is undefined.
x≤0
Step 6.2
The equation is undefined where the denominator equals 0, the argument of a square root is less than 0, or the argument of a logarithm is less than or equal to 0.
x≤0
(-∞,0]
x≤0
(-∞,0]
Step 7
Critical points to evaluate.
x=1e14
Step 8
Evaluate the second derivative at x=1e14. If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.
7(1e14)2+12(1e14)2ln(1e14)
Step 9
Step 9.1
Simplify each term.
Step 9.1.1
Apply the product rule to 1e14.
712(e14)2+12(1e14)2ln(1e14)
Step 9.1.2
One to any power is one.
71(e14)2+12(1e14)2ln(1e14)
Step 9.1.3
Multiply the exponents in (e14)2.
Step 9.1.3.1
Apply the power rule and multiply exponents, (am)n=amn.
71e14⋅2+12(1e14)2ln(1e14)
Step 9.1.3.2
Cancel the common factor of 2.
Step 9.1.3.2.1
Factor 2 out of 4.
71e12(2)⋅2+12(1e14)2ln(1e14)
Step 9.1.3.2.2
Cancel the common factor.
71e12⋅2⋅2+12(1e14)2ln(1e14)
Step 9.1.3.2.3
Rewrite the expression.
71e12+12(1e14)2ln(1e14)
71e12+12(1e14)2ln(1e14)
71e12+12(1e14)2ln(1e14)
Step 9.1.4
Combine 7 and 1e12.
7e12+12(1e14)2ln(1e14)
Step 9.1.5
Apply the product rule to 1e14.
7e12+1212(e14)2ln(1e14)
Step 9.1.6
One to any power is one.
7e12+121(e14)2ln(1e14)
Step 9.1.7
Multiply the exponents in (e14)2.
Step 9.1.7.1
Apply the power rule and multiply exponents, (am)n=amn.
7e12+121e14⋅2ln(1e14)
Step 9.1.7.2
Cancel the common factor of 2.
Step 9.1.7.2.1
Factor 2 out of 4.
7e12+121e12(2)⋅2ln(1e14)
Step 9.1.7.2.2
Cancel the common factor.
7e12+121e12⋅2⋅2ln(1e14)
Step 9.1.7.2.3
Rewrite the expression.
7e12+121e12ln(1e14)
7e12+121e12ln(1e14)
7e12+121e12ln(1e14)
Step 9.1.8
Combine 12 and 1e12.
7e12+12e12ln(1e14)
Step 9.1.9
Move e14 to the numerator using the negative exponent rule 1bn=b-n.
7e12+12e12ln(e-14)
Step 9.1.10
Expand ln(e-14) by moving -14 outside the logarithm.
7e12+12e12(-14ln(e))
Step 9.1.11
The natural logarithm of e is 1.
7e12+12e12(-14⋅1)
Step 9.1.12
Multiply -1 by 1.
7e12+12e12(-14)
Step 9.1.13
Cancel the common factor of 4.
Step 9.1.13.1
Move the leading negative in -14 into the numerator.
7e12+12e12⋅-14
Step 9.1.13.2
Factor 4 out of 12.
7e12+4(3)e12⋅-14
Step 9.1.13.3
Cancel the common factor.
7e12+4⋅3e12⋅-14
Step 9.1.13.4
Rewrite the expression.
7e12+3e12⋅-1
7e12+3e12⋅-1
Step 9.1.14
Combine 3e12 and -1.
7e12+3⋅-1e12
Step 9.1.15
Multiply 3 by -1.
7e12+-3e12
Step 9.1.16
Move the negative in front of the fraction.
7e12-3e12
7e12-3e12
Step 9.2
Combine fractions.
Step 9.2.1
Combine the numerators over the common denominator.
7-3e12
Step 9.2.2
Subtract 3 from 7.
4e12
4e12
4e12
Step 10
x=1e14 is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.
x=1e14 is a local minimum
Step 11
Step 11.1
Replace the variable x with 1e14 in the expression.
f(1e14)=(1e14)4ln(1e14)
Step 11.2
Simplify the result.
Step 11.2.1
Simplify the expression.
Step 11.2.1.1
Apply the product rule to 1e14.
f(1e14)=14(e14)4⋅ln(1e14)
Step 11.2.1.2
One to any power is one.
f(1e14)=1(e14)4⋅ln(1e14)
f(1e14)=1(e14)4⋅ln(1e14)
Step 11.2.2
Simplify the denominator.
Step 11.2.2.1
Multiply the exponents in (e14)4.
Step 11.2.2.1.1
Apply the power rule and multiply exponents, (am)n=amn.
f(1e14)=1e14⋅4⋅ln(1e14)
Step 11.2.2.1.2
Cancel the common factor of 4.
Step 11.2.2.1.2.1
Cancel the common factor.
f(1e14)=1e14⋅4⋅ln(1e14)
Step 11.2.2.1.2.2
Rewrite the expression.
f(1e14)=1e⋅ln(1e14)
f(1e14)=1e⋅ln(1e14)
f(1e14)=1e⋅ln(1e14)
Step 11.2.2.2
Simplify.
f(1e14)=1e⋅ln(1e14)
f(1e14)=1e⋅ln(1e14)
Step 11.2.3
Move e14 to the numerator using the negative exponent rule 1bn=b-n.
f(1e14)=1e⋅ln(e-14)
Step 11.2.4
Expand ln(e-14) by moving -14 outside the logarithm.
f(1e14)=1e⋅(-14⋅ln(e))
Step 11.2.5
The natural logarithm of e is 1.
f(1e14)=1e⋅(-14⋅1)
Step 11.2.6
Multiply -1 by 1.
f(1e14)=1e⋅(-14)
Step 11.2.7
Multiply 1e by 14.
f(1e14)=-1e⋅4
Step 11.2.8
Move 4 to the left of e.
f(1e14)=-14e
Step 11.2.9
The final answer is -14e.
y=-14e
y=-14e
y=-14e
Step 12
These are the local extrema for f(x)=x4ln(x).
(1e14,-14e) is a local minima
Step 13