Question Statement

Show that:

$\frac{d^2}{d{x}^2}(fg)=f^{\prime \prime }g+2f^{\prime }{g}^{\prime }+f{g}^{\prime \prime }$

$\frac{d^3}{d{x}^3}(fg)=f^{\prime \prime }g+3f^{\prime \prime }{g}^{\prime }+3f^{\prime }{g}^{\prime \prime }+f{g}^{\prime \prime }$

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Background and Explanation

This problem demonstrates Leibniz's rule for higher-order derivatives of a product, which follows a pattern similar to the binomial theorem. You need to apply the product rule repeatedly, treating each derivative operation as a new differentiation task.

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Solution

Finding the Second Derivative

We begin by finding the first derivative of the product $fg$ using the product rule:

$\frac{d}{dx}(fg)=\frac{d}{dx}(f)\cdot g+f\cdot \frac{d}{dx}(g)=f^{\prime }g+f{g}^{\prime }$

Now, differentiate this result with respect to $x$ again to obtain the second derivative:

$\frac{d^2}{d{x}^2}(fg)=\frac{d}{dx}\left(f^{\prime }g\right)+\frac{d}{dx}\left(f{g}^{\prime }\right)$

Apply the product rule to each term separately:

$=\left[\frac{d}{dx}\left(f^{\prime }\right)\cdot g+f^{\prime }\cdot \frac{d}{dx}(g)\right]+\left[\frac{d}{dx}(f)\cdot g^{\prime }+f\cdot \frac{d}{dx}\left(g^{\prime }\right)\right]$

$=f^{\prime \prime }g+f^{\prime }{g}^{\prime }+f^{\prime }{g}^{\prime }+f{g}^{\prime \prime }$

Combine like terms:

$\frac{d^2}{d{x}^2}(fg)=f^{\prime \prime }g+2f^{\prime }{g}^{\prime }+f{g}^{\prime \prime }$

Finding the Third Derivative

Now differentiate the second derivative result with respect to $x$:

$\frac{d^3}{d{x}^3}(fg)=\frac{d}{dx}\left(f^{\prime \prime }g\right)+2\frac{d}{dx}\left(f^{\prime }{g}^{\prime }\right)+\frac{d}{dx}\left(f{g}^{\prime \prime }\right)$

Apply the product rule to each of the three terms:

$=\left[\frac{d}{dx}\left(f^{\prime \prime }\right)\cdot g+f^{\prime \prime }\cdot \frac{d}{dx}(g)\right]+2\left[\frac{d}{dx}\left(f^{\prime }\right)\cdot g^{\prime }+f^{\prime }\cdot \frac{d}{dx}\left(g^{\prime }\right)\right]+\left[\frac{d}{dx}(f)\cdot g^{\prime \prime }+f\cdot \frac{d}{dx}\left(g^{\prime \prime }\right)\right]$

Simplify each derivative:

$=f^{\prime \prime }g+f^{\prime \prime }{g}^{\prime }+2\left(f^{\prime \prime }{g}^{\prime }+f^{\prime }{g}^{\prime \prime }\right)+f^{\prime }{g}^{\prime \prime }+f{g}^{\prime \prime }$

Expand the terms:

$=f^{\prime \prime }g+f^{\prime \prime }{g}^{\prime }+2f^{\prime \prime }{g}^{\prime }+2f^{\prime }{g}^{\prime \prime }+f^{\prime }{g}^{\prime \prime }+f{g}^{\prime \prime }$

Hence proved.

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$
• Leibniz Rule for Higher Derivatives: $\frac{d^n}{d{x}^n}(fg)={\sum }_{k=0}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)f^{(n-k)}{g}^{(k)}$
• Successive Differentiation: Repeated application of differentiation rules to higher-order derivatives

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Summary of Steps

1. First Derivative: Apply product rule to get $\frac{d}{dx}(fg)=f^{\prime }g+f{g}^{\prime }$
2. Second Derivative: Differentiate the result from step 1, applying product rule to both terms $f^{\prime }g$ and $f{g}^{\prime }$
3. Simplify: Collect like terms ($f^{\prime }{g}^{\prime }+f^{\prime }{g}^{\prime }=2f^{\prime }{g}^{\prime }$) to obtain $f^{\prime \prime }g+2f^{\prime }{g}^{\prime }+f{g}^{\prime \prime }$
4. Third Derivative: Differentiate the second derivative result, treating it as three terms: $f^{\prime \prime }g$, $2f^{\prime }{g}^{\prime }$, and $f{g}^{\prime \prime }$
5. Apply Product Rule: Differentiate each term using the product rule (note: constant coefficients remain)
6. Expand and Collect: Expand all brackets and group like terms to verify the pattern matches the required formula