Problem 33

Question

Find the second derivative of the product $u(x) v(x)$. Find the third derivative. Test your formulas on $u=v=x$.

Step-by-Step Solution

Verified

Answer

The second derivative is $u''v + 2u'v' + uv''$, and the third is $u'''v + 3u''v' + 3u'v'' + uv'''$. For $u=v=x$, the second derivative is 2, and the third is 0.

1Step 1: Apply the Product Rule First Derivative

To find the derivative of the product of two functions, use the product rule, which states $(uv)' = u'v + uv'$. So, the first derivative of $u(x) v(x)$ is:\[(uv)' = u'v + uv'\].

2Step 2: Differentiate Again for Second Derivative

Differentiate the expression from Step 1 using the product rule again. For the second derivative, differentiate $(u'v + uv')$ with respect to $x$:\[(uv)'' = (u'v)' + (uv')' = (u''v + u'v') + (u'v' + uv'')\]Simplifying, we have:\[(uv)'' = u''v + 2u'v' + uv''\].

3Step 3: Differentiate Once More for Third Derivative

Differentiate the second derivative expression to obtain the third derivative:\[(uv)''' = (u''v + 2u'v' + uv'')'\]Use the product rule once more for each term:\[(uv)''' = (u''v)' + (2u'v')' + (uv'')'\]Simplifying gives:\[(uv)''' = u'''v + 2u''v' + u'v'' + 2(u''v' + u'v'') + (u'v'' + uv''')\]Final simplification:\[(uv)''' = u'''v + 3u''v' + 3u'v'' + uv'''\].

4Step 4: Verify with u=v=x

Substitute $u=v=x$ into the formulas. The derivatives of $x$ are: $u' = 1$, $u'' = 0$, $u''' = 0$; likewise for $v$.Apply these to the second derivative formula:\[(uv)'' = (x)'' = 0 \cdot x + 2 \cdot 1 \cdot 1 + x \cdot 0 = 2\]Apply these to the third derivative formula:\[(uv)''' = (x)''' = 0 \cdot x + 3 \cdot 0 \cdot 1 + 3 \cdot 1 \cdot 0 + x \cdot 0 = 0\].

Key Concepts

Product ruleSecond derivativeThird derivativeFunction derivatives

Product rule

The Product Rule is an essential technique in calculus that helps us find the derivative of a product of two functions. When we have two functions, say $ u(x) $ and $ v(x) $, the product rule allows us to differentiate their product $ y = u(x) \cdot v(x) $. The formula for the product rule is:

$ (uv)' = u'v + uv' $

In this rule:

$ u' $ is the derivative of $ u(x) $
$ v' $ is the derivative of $ v(x) $

Using the product rule is like distributing the derivative over the product, but requires adjusting by swapping derivatives between the two functions.
It's a foundational skill in calculus differentiation and understanding this rule is crucial for tackling more complex differentiation problems.

Second derivative

After finding the first derivative using the product rule, the next step involves computing the second derivative. The second derivative is essentially the derivative of the derivative, denoted as $ (uv)'' $. It provides insights into the curvature or the concavity of the function.
To find the second derivative of $ uv $, we apply the product rule again to each component of the first derivative $(uv)' = u'v + uv'$. The steps are as follows:

Differentiating $ u'v $ gives $ u''v + u'v' $
Differentiating $ uv' $ gives $ u'v' + uv'' $

Combine these results to obtain the second derivative:

$ (uv)'' = u''v + 2u'v' + uv'' $

This helps determine how sharply the function is curving at a particular point.

Third derivative

The third derivative, denoted as $(uv)'''$, goes another step beyond the second.
It represents the rate of change of the curvature, helping us understand changes in the function's "bending" behavior.
To compute the third derivative, we need to differentiate the second derivative $(uv)'' = u''v + 2u'v' + uv''$ yet again. Here's how:

Differentiate $ u''v $ to get $ u'''v + u''v' $
Differentiate $ 2u'v' $ to get $ 2(u''v' + u'v'') $
Differentiate $ uv'' $ to get $ u'v'' + uv''' $

Put these together for the third derivative:

$ (uv)''' = u'''v + 3u''v' + 3u'v'' + uv''' $

The third derivative can clue you into points of inflection where the nature of the curvature changes.

Function derivatives

Function derivatives are tools that help us understand how functions behave.
Derivatives tell us the rate at which a function is changing at any given point.
In calculus, this concept extends beyond just simple functions and applies to combinations of functions.The derivative of a function $ f(x) $ not only shows us its slope but also its instantaneous rate of change.
For complex functions involving products or compositions, we use rules like the product rule, chain rule, or quotient rule to find derivatives.
With each successive derivative (first, second, third, and beyond), we get deeper insights:

First derivative $ f'(x) $: Shows rate of change and tangent slope
Second derivative $ f''(x) $: Informs curvature, concavity, convexity
Third derivative $ f'''(x) $: Relates to changes in curvature, getting into more refined behavior

By mastering derivatives, students can solve challenges involving motion, optimization, and more.

Problem 33

Other exercises in this chapter

Problem 33

For the polynomial $f(x)=2 x-5 x^{2}+7 x^{3}$ find (a) $\lim _{x \rightarrow 1} f(x)$ (b) $\lim _{x \rightarrow \infty} f(x)$ (c) \(\lim _{x \rightarrow \

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Problem 33

If $f(0)=0$ and $f^{\prime}(0)=3$, rank these functions from smallest to largest as $x$ decreases to zero: $$ f(x), x, x f(x), f(x)+2 x, 2(f(x)-x), \quad

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Problem 33

Explain the second derivative in your own words.

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Problem 34

Suppose $d f / d x=x^{-2}-x^{-3}$. Find $f(x)$.

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