Problem 30

Question

Differentiate $$ f(x)=(a-x)(a+x) $$ with respect to $x$. Assume that $a$ is a positive constant.

Step-by-Step Solution

Verified

Answer

The derivative of $ f(x) = (a-x)(a+x) $ is $-2x$.

1Step 1: Expand the Function

The function given is $ f(x) = (a-x)(a+x) $. To differentiate this, it is simpler to first expand it. This is a difference of squares:\[f(x) = a^2 - x^2\]

2Step 2: Apply the Power Rule

Now we differentiate each term in the expression $ f(x) = a^2 - x^2 $ term by term. The derivative of a constant is 0, and the derivative of $-x^2$ is found using the power rule.The power rule states: If $ f(x) = x^n $, then $ f'(x) = nx^{n-1} $.So, the derivative of $-x^2$ is:\[-2x\]

3Step 3: Write the Final Derivative

Combining the results, the derivative of $ f(x) = a^2 - x^2 $ is:\[f'(x) = 0 - 2x = -2x\]

Key Concepts

Power RuleConstantDifference of Squares

Power Rule

The power rule is a fundamental principle in calculus for finding derivatives. It simplifies the process of differentiation by providing a straightforward way to differentiate polynomials. When applying the power rule, you follow this simple formula: if you have a function of the form $f(x) = x^n$, the derivative $f'(x)$ is $nx^{n-1}$. In essence,

You bring the exponent down in front as a coefficient.
Then, you reduce the exponent by one.

For example, when differentiating $-x^2$, you multiply the power (2) by the constant coefficient (-1), obtaining -2. Then, decrease the power of $x$ from 2 to 1, resulting in $-2x$.
This method is especially helpful when dealing with polynomial functions. It saves time and reduces errors, providing an efficient way of handling differentiation tasks.

Constant

In calculus, a constant is a term in a function that does not change. It is typically represented by a number or a symbol, like $a$ in our exercise. When differentiating a function, remember that the derivative of a constant term is always zero.

No matter what the constant value is, it does not depend on the variable $x$, so its rate of change is zero.
This is why, in our exercise, when differentiating $a^2$, we simply get 0.

Understanding this rule is crucial because it helps in focusing on terms involving the variable $x$ when finding derivatives.
This means you can immediately drop any constant terms from your differentiation calculations, making the process quicker and more efficient.

Difference of Squares

The difference of squares is a special algebraic pattern that makes expanding expressions quicker and easier. It follows the formula: $(a+b)(a-b) = a^2 - b^2$. This means, whenever you have a product of two binomials differing only by their signs, you can expand it to a subtraction of squares.
In our exercise, we used the difference of squares to simplify the given function $(a-x)(a+x)$. Applying this formula, we see:

If $a^2$ is the square of the term $a$.
And $x^2$ is the square of $x$.
The expression simplifies to $a^2 - x^2$.

By recognizing the difference of squares early on, we save time and simplify the subsequent differentiation process. This concept is broad-reaching, appearing frequently in algebra and calculus, so being comfortable with it can greatly aid in tackling similar mathematical problems.

Problem 30

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