Calculus Examples

Find the second derivative.
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Find the first derivative.
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By the Sum Rule, the derivative of with respect to is .
Evaluate .
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Since is constant with respect to , the derivative of with respect to is .
Differentiate using the Power Rule which states that is where .
Multiply by .
Evaluate .
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Since is constant with respect to , the derivative of with respect to is .
Differentiate using the Power Rule which states that is where .
Multiply by .
Find the second derivative.
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By the Sum Rule, the derivative of with respect to is .
Evaluate .
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Since is constant with respect to , the derivative of with respect to is .
Differentiate using the Power Rule which states that is where .
Multiply by .
Evaluate .
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Since is constant with respect to , the derivative of with respect to is .
Differentiate using the Power Rule which states that is where .
Multiply by .
The second derivative of with respect to is .
Set the second derivative equal to then solve the equation .
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Set the second derivative equal to .
Add to both sides of the equation.
Divide each term by and simplify.
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Divide each term in by .
Cancel the common factor of .
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Cancel the common factor.
Divide by .
Cancel the common factor of and .
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Factor out of .
Cancel the common factors.
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Factor out of .
Cancel the common factor.
Rewrite the expression.
Find the points where the second derivative is .
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Substitute in to find the value of .
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Replace the variable with in the expression.
Simplify the result.
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Simplify each term.
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Apply the product rule to .
One to any power is one.
Raise to the power of .
Combine and .
Apply the product rule to .
One to any power is one.
Raise to the power of .
Combine and .
Move the negative in front of the fraction.
To write as a fraction with a common denominator, multiply by .
Write each expression with a common denominator of , by multiplying each by an appropriate factor of .
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Multiply and .
Multiply by .
Combine the numerators over the common denominator.
Simplify the numerator.
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Multiply by .
Subtract from .
Move the negative in front of the fraction.
The final answer is .
The point found by substituting in is . This point can be an inflection point.
Split into intervals around the points that could potentially be inflection points.
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Replace the variable with in the expression.
Simplify the result.
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Multiply by .
Subtract from .
The final answer is .
At , the second derivative is . Since this is negative, the second derivative is decreasing on the interval
Decreasing on since
Decreasing on since
Substitute a value from the interval into the second derivative to determine if it is increasing or decreasing.
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Replace the variable with in the expression.
Simplify the result.
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Multiply by .
Subtract from .
The final answer is .
At , the second derivative is . Since this is positive, the second derivative is increasing on the interval .
Increasing on since
Increasing on since
An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is .
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