Reciprocals

You can use the quotient rule \( \frac{u’v - v’u}{v^2} \) to compute the derivative of a reciprocal.

A reciprocal function is in the form \( \frac{1}{f} \). So for this \( u = 1 \) and \( v = f \).

Before substituting into the quotient rule form we find the derivative of \( u \) which for \( 1 \) is \( 0 \) because the derivative of any constant is zero. \(v \) is just a general function \( f \) so this just \( f’ \).

Substituting we get

\[ \frac{0 \cdot v - 1 \cdot v’}{v^2} \]

\[ \frac{-v’}{v^2} \]

We can then bring the \( v^2 \) divisor up as a multiplier by raising it to the negative power so that we have

\[ -v^{-2}v’ \]

This gives a general formula for computing the derivatives of reciprocal functions.

nx case

The general formula \( -v^{-2}v’ \) can be applied to derivatives in the \( x^n \) format to yield insights about the rule for taking derivatives of an \( x^n \) function.

By substituing into the reciprocal formula we see that

\[ \frac{d}{dx} \frac{1}{x^n} = -(x^n)^{-2}(nx^{n-1}) \]

Then multiplying out indices is the first term

\[ = -x^{-2n}(nx^{n-1}) \]

Using the additive rules that apply when muiltiplying indices we know that \( x^{-2n} \cdot x^{n-1} \) is \( -2n + n-1 \) and \( -2n + n = -n \) so the indices in this expression resolve to \( -n-1 \).

So, evaluating the whole expression we get

\[ = -nx^{-n-1} \]

The general \( \frac{d}{dx} x^n \) formula is \( nx^{n-1} \) and as \( \frac{1}{x^n} = x^{-n} \) we can show that formula holds true even for negative powers of \( n \) because if \( n \) is negative it simply resolves to \( -nx^{-n-1} \) and by applying the quotient rule to generaly reciprocals and then the \( x^n \) formula we have shown why.