How Reciprocal and Chain Rules Differentiate 1/(x+a)

A reciprocal of a linear expression can be differentiated efficiently by rewriting it with a negative exponent. If

$$g(x)=\frac{1}{u(x)}=[u(x)]^{-1},$$

then the power rule and chain rule give

$$g'(x)=-[u(x)]^{-2}u'(x) =-\frac{u'(x)}{[u(x)]^2}.$$

Rewrite reciprocals before differentiating

For a linear denominator $u(x)=mx+b$,

$$\frac{1}{mx+b}=(mx+b)^{-1}.$$

Differentiating gives

$$\frac{d}{dx}(mx+b)^{-1} =-(mx+b)^{-2}\cdot m =-\frac{m}{(mx+b)^2}.$$

The negative sign comes from differentiating the exponent $-1$, while the factor $m$ comes from differentiating the inside function.

The domain restriction survives differentiation

A reciprocal function is undefined wherever its denominator is zero. If $mx+b=0$, then the original function is not defined at that input, so a derivative of that function cannot be assigned there by the ordinary differentiation rules.

For example, if

$$r(x)=\frac{1}{2x+5},$$

then

$$r'(x)=-\frac{2}{(2x+5)^2},$$

with $x\ne-5/2$.

Quotient rule and negative-exponent methods agree

The same reciprocal can also be treated as a quotient with numerator $1$. The quotient rule produces the same derivative. Rewriting with a negative exponent is usually shorter when the numerator is constant, while the quotient rule is useful when both numerator and denominator vary.