Problem 92 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) is a differentiable function. Find the derivative of the reciprocal function \(g(x)=1 / f(x)\) at those points \(x\) where \(f(x) \neq 0\)

Step-by-Step Solution

Answer

The derivative is \(g'(x) = \frac{-f'(x)}{(f(x))^2}.\)

Step 1: Understand the Problem

We need the derivative of the function \(g(x) = \frac{1}{f(x)}\) where \(f(x)\) is differentiable and \(f(x) eq 0\). This involves finding \(g'(x)\).

Step 2: Use the Quotient Rule

Recall the quotient rule: if you have a quotient \(g(x) = \frac{u(x)}{v(x)}\), then its derivative is given by \(g'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\). In our case, \(u(x) = 1\) and \(v(x) = f(x)\).

Step 3: Apply the Quotient Rule

Since \(u(x) = 1\) and \(v(x) = f(x)\), the derivative becomes:\[g'(x) = \frac{f(x) \cdot 0 - 1 \cdot f'(x)}{(f(x))^2} = \frac{-f'(x)}{(f(x))^2}.\]

Step 4: Write the Final Expression

Thus, the derivative of \(g(x) = \frac{1}{f(x)}\) at points where \(f(x) eq 0\) is given by:\[g'(x) = \frac{-f'(x)}{(f(x))^2}.\]

Key Concepts

Differentiable Function

When we say a function is differentiable, it means that the function has a derivative at every point in its domain. A differentiable function is smooth and continuous; it doesn't have any sharp corners or jumps. Most functions that you come across early in calculus, like polynomials, sine, cosine, or exponential functions, are differentiable over their entire domain. For instance, if we consider a function like \(f(x)\), being differentiable at a point \(x=a\) implies that we can find the derivative \(f'(a)\). This tells us the slope of the tangent line to the curve at that point.

• A function must be continuous to be differentiable, but not every continuous function is differentiable.
• If a function is not differentiable at a point, it might be because there's a corner, cusp, vertical tangent, or discontinuity at that point.

Understanding differentiability is crucial in calculus because it sets the stage for discussing derivatives and their application across various problems.

Quotient Rule

The quotient rule is a method for finding the derivative of a function that is the quotient of two differentiable functions. This is particularly useful when dealing with functions expressed as \(\frac{u(x)}{v(x)}\), where both \(u(x)\) and \(v(x)\) are differentiable functions, and \(v(x) eq 0\).The quotient rule formula is:\[g'(x) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}\]This formula helps compute the derivative of the entire quotient instead of differentiating each part separately. Some key points about using the Quotient Rule:

• The order in which you subtract \(u(x) \cdot v'(x)\) from \(v(x) \cdot u'(x)\) is crucial. Always subtract in this order.
• Ensure that the denominator \((v(x))^2\) is never zero as this would make the function undefined.

In our specific problem, since \(g(x) = \frac{1}{f(x)}\), we applied the quotient rule with \(u(x) = 1\) and \(v(x) = f(x)\), resulting in \(g'(x) = \frac{-f'(x)}{(f(x))^2}\). This shows how the rule neatly simplifies even when dealing with a reciprocal.

Derivative Computation

Derivative computation involves using rules and formulas to find the derivative of a given function. In calculus, this is usually about finding the rate at which something changes. The most common rules for finding derivatives include the power rule, product rule, chain rule, and quotient rule.For our given problem, after identifying \(g(x) = \frac{1}{f(x)}\), derivative computation involved these steps:

• Setting \(u(x) = 1\) and \(v(x) = f(x)\).
• Applying the quotient rule to these specific functions.
• Finding the derivatives \(u'(x) = 0\) and \(f'(x)\).
• Substituting into the formula to get \(g'(x) = \frac{-f'(x)}{(f(x))^2}\).

Such calculations may seem complex at first but with practice, you can become proficient. It is important to clearly follow each step and ensure the function remains defined at every point you are evaluating, especially for fractions where the denominator must not be zero.