Question

Proof Let \(u\) be a differentiable function of \(x .\) Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\)

Short answer

\(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0\)

Step-by-step solution

Apply the Chain Rule

The chain rule states that \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). Here, \(f(u) = \sqrt{u}\) and \(g(u) = u^2\). Applying the chain rule, we get \[ \frac{d}{dx} |u| = \frac{1}{2\sqrt{u^2}} \cdot 2u \cdot u' \] which simplifies to \( \frac{u \cdot u'}{|u|}\). Note that \(|u| = \sqrt{u^2}\) will always be positive for \( u\neq 0 \).

Convert the Absolute Expression to Fraction Form

We notice that the numerator \( u \cdot u' \) is just the product of \( u \) (itself) and its derivative. The absolute operation in the denominator can be expressed as \( u/|u| \) because that always will return the sign of \( u \). Hence, \[ \frac{u \cdot u'}{|u|} = u' \cdot \frac{u}{|u|} \] which is the required form.

Key concepts

Chain Rule

The chain rule is essential in calculus for finding the derivative of composite functions. When a function is a composition of two or more functions, such as \( f(g(x)) \), the chain rule provides a method to differentiate it.

So, if you have a function \( f \) that is dependent on \( g \), which in turn is dependent on \( x \), the derivative of \( f \) with respect to \( x \) is the derivative of \( f \) with respect to \( g \) times the derivative of \( g \) with respect to \( x \). Symbolically, it's written as \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \].

For a clearer understanding, imagine \( g(x) \) as an intermediate step; it's the inside function that we first differentiate, and then we multiply by the derivative of the outside function, \( f \), which has been applied to the result of \( g(x) \). This rule is a powerful tool because it simplifies the differentiation process for complex expressions.

Absolute Value Differentiation

Differentiation of absolute values requires careful consideration, as it involves piecewise functions due to the nature of absolute value. The absolute value of a function \( u \), denoted as \( |u| \), represents the distance of \( u \) on the number line from zero, without considering the direction (positive or negative).

To differentiate an absolute value function, one common strategy is to express it as \( |u| = \sqrt{u^2} \), which allows the use of other differentiation rules. In the given exercise, we take advantage of this by differentiating \( \sqrt{u^2} \) instead of \( |u| \) directly. Once we apply the chain rule, as shown in the step-by-step solution provided, we gain an expression that handles the sign of \( u \) neatly in the form of \( \frac{u}{|u|} \).

It's crucial to remember that \( |u| \) can't be negative, thus when dealing with the derivative, if \( u \) is not zero, the differentiation works out to the form shown in the solution. However, the absolute value function is not differentiable at zero since it has a sharp corner there, which breaks the smoothness required for differentiation.

Derivative of a Function

The derivative of a function represents the rate at which the function's output values are changing with respect to changes in the input values. To put it simply, it measures how much the function \( f(x) \) increases or decreases as \( x \) increases by an infinitesimally small amount.

The derivative of a function at a point is the slope of the tangent line to the function's graph at that point. Graphically, it shows us the steepness and direction of the function curve at a particular instant. If \( f'(x) \) is positive, the function is increasing at \( x \), and if \( f'(x) \) is negative, the function is decreasing at \( x \).

In practical terms, we often use rules and formulas, such as the product rule, quotient rule, chain rule, power rule, and others, to calculate the derivative of different kinds of functions. These tools make finding derivatives a more systematic process rather than a matter of guesswork.