Rational Functions With Linear Denominators

Why Rewrite Before Differentiating?

A rational function with a linear denominator may look like a single complicated fraction. If its numerator is also linear, however, the function can often be rewritten as a constant plus a reciprocal term. That form makes the derivative straightforward to find using the reciprocal formula and the sum rule.

For example, the numerator \(x+b\) differs from the denominator \(x-a\) by the constant \(a+b\):

Dividing by \(x-a\), wherever \(x-a\ne 0\), gives a constant plus a reciprocal linear term. The denominator restriction matters throughout: the original function is not defined at \(x=a\), and rewriting it does not change that domain.

The Rewrite and Its Domain

Let \(a\) and \(b\) be constants, and consider

The denominator is zero at \(x=a\), so the domain is all real \(x\) except \(a\). Using \(x+b=(x-a)+(a+b)\), we obtain the identity

This is an identity on the domain of the original function. It does not say that the function is defined at the excluded input \(x=a\).

The same algebra works when the numerator has a coefficient on \(x\). For constants \(m,n,a\),

so, for \(x\ne a\),

The coefficient \(ma+n\) is the constant remainder after expressing the numerator as a multiple of the denominator plus a constant.

Differentiate the Rewritten Form

The reciprocal formula for a shifted reciprocal is

The sum rule says that the derivative of a sum of differentiable functions is the sum of their derivatives. A constant has derivative zero, and a constant multiple may be taken outside a derivative. Applying these facts to the rewrite gives

For the more general numerator \(mx+n\), the same reasoning yields

Why the Derivative Formula Works

The algebraic rewrite proves the decomposition. The reciprocal formula explains the derivative of the nonconstant term. To see directly why the reciprocal formula applies, fix \(x\ne a\). For nonzero \(h\) small enough that \(x+h\ne a\), the difference quotient for \(g(x)=1/(x-a)\) is

As \(h\) approaches zero, \(x+h-a\) approaches \(x-a\), which is nonzero because \(x\ne a\). Therefore the limit exists and equals

Multiplying \(g\) by the constant \(a+b\) multiplies its derivative by the same constant. Adding the constant \(1\) does not affect the derivative. These facts give the stated formula for \(f\), at every point where \(f\) is defined.

Worked Examples

A Cancellation Case and Its Domain

If \(a+b=0\), the reciprocal term in the decomposition has coefficient zero. For example, when \(a=3\) and \(b=-3\),

On its domain, this function is constant, so its derivative is zero for every \(x\ne 3\). The original expression is still undefined at \(x=3\). Differentiation here concerns inputs in the function's domain; the rewrite does not authorize substituting the excluded input into the original function.

Common Pitfall: Getting the Remainder or Sign Wrong

A frequent mistake is to use \(b\), rather than \(a+b\), as the remainder in the decomposition. For \((x+5)/(x-2)\), that error would give

Indeed, multiplying the proposed right-hand side by \(x-2\) produces \((x-2)+5=x+3\), not \(x+5\). The correct remainder is \(a+b=2+5=7\), because

Another sign error is to forget the minus sign in the reciprocal formula. Since

the derivative of \(1+7/(x-2)\) is \(-7/(x-2)^2\), not \(7/(x-2)^2\). Squaring the denominator does not remove the negative sign produced by differentiating the reciprocal.

A Reliable Procedure

Why This Form Is Useful

The rewrite separates two different parts of the function: a constant term and a shifted reciprocal term. The constant contributes zero to the derivative, while all of the derivative comes from the reciprocal term. This makes the source of both the coefficient and the negative sign visible.

It also provides a dependable check on the algebra. Multiplying the proposed decomposition by the original denominator must recover the numerator. If it does not, the remainder or the constant term is incorrect. Keeping the excluded input visible at each stage prevents a different error: treating an algebraically simplified expression as though it were defined at every real number.