# Algebra of Derivatives

After we learned about limits, we quickly had
to find out whether the limit of a sum was the
sum of limits, and so on. Likewise, it would make
our lives much easier if the derivative
of a sum is the sum of the derivatives. It is.

[Theorem: [DerivativeAddition]] Suppose that \(f\) and \(g\) are functions whose
derivatives exist at a point \(x_0\). Then
the derivative of \(f+g\) exists at \(x_0\) and
\[(f+g)'(x_0)=f'(x_0)+g'(x_0).\]

This is easy to see from the properties of limits.
In particular,

[LimitAddition] shows that
\begin{align*}
f'(x_0)+g'(x_0)&=
\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}+\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}\\
&= \lim_{x\to x_0}\frac{f(x)+g(x)-f(x_0)-g(x_0)}{x-x_0}\\
&= \lim_{x\to x_0}\frac{(f+g)(x)-(f+g)(x_0)}{x-x_0}.\\
\end{align*}
Since the quantity on the left exists, so must the quantity on the right.

The derivative of a product of functions is only a little more tricky.
Suppose that \(f\) and \(g\) are functions whose
derivatives exist at a point \(x_0\). Then
[LimitProduct] shows that
\begin{align*}
\lim_{x\to x_0}\frac{fg(x)-fg(x_0)}{x-x_0}
&= \lim_{x\to x_0}\frac{f(x)g(x)-f(x_0)g(x_0)}{x-x_0}\\
&= \lim_{x\to x_0}\frac{f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)}{x-x_0}.\\
&= \lim_{x\to x_0}f(x)\frac{g(x)-g(x_0)}{x-x_0}+\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}g(x_0).\\
&= f(x_0)g'(x_0)+f'(x_0)g(x_0).
\end{align*}
This establishes the following result.

[Theorem: [ProductRule]] Suppose that \(f\) and \(g\) are functions whose
derivatives exist at a point \(x_0\). Then the
derivative of the product \(fg\) exists at \(x_0\) and
\[(fg)'(x_0) = f(x_0)g'(x_0)+f'(x_0)g(x_0).\]

At this point we can differentiate sums and products of
functions. If we could find the derivative of the reciprocal
of a differentiable function, then we would be able to find
derivatives of a large collection of interesting functions.
Consider a reciprocal for a function \(f\) that has
a derivative at \(x_0\), where we assume that
\(f(x_0)\ne 0\). We can cross-multiply to find that
\begin{align*}
\lim_{x\to x_0}\frac{ \frac1{f(x)}-\frac1{f(x_0)} }{x-x_0}
&= \lim_{x\to x_0}\frac{f(x_0)-f(x)}{f(x)f(x_0)(x-x_0)}\\
&= \lim_{x\to x_0}\frac1{f(x)f(x_0)}\lim_{x\to x_0}\frac{f(x_0)-f(x)}{x-x_0}\\
&= -\frac{f'(x_0)}{f(x_0)^2}.
\end{align*}
This gives the next result.

[Theorem: [DerivativeReciprocal]] Suppose that \(f\) is differentiable at \(x_0\), and
that \(f(x_0)\ne0\). Then the derivative of
\(1/f\) exists at \(x_0\), and is given by
\[\left(\frac1f\right)'(x_0) = \frac{f'(x_0)}{f(x_0)^2}.\]

This, together with [ProductRule]quickly gives another useful result.

[Theorem: [QuotientRule]] Suppose that \(f\) and \(g\) are both differentiable
at \(x_0\), and that \(f(x_0)\ne 0\). Then
\(g/f\) is differentiable at \(x_0\) and
\[\left(\frac{g}f\right)'(x_0)=\frac{g'(x_0)f(x_0)-f'(x_0)g(x_0)}{f(x_0)^2}\]

To see why this is, we just apply

[ProductRule] and

[DerivativeReciprocal].
\begin{align*}
\left(\frac{g}f\right)'(x_0)
&= \left(\frac1f\right)(x_0)g'(x_0)+\left(\frac1f\right)'(x_0)g(x_0)\\
&= \frac{g'(x_0)}{f(x_0)}-\frac{f'(x_0)}{f(x_0)^2}g(x_0)\\
&= \frac{g'(x_0)f(x_0)-f'(x_0)g(x_0)}{f(x_0)^2}.
\end{align*}