# 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*}