Differentiating Linear Functions
Differentiating Products
UBC Calculus Online Course Notes

Differentiating Sums of Functions

Many functions are built by summing two simpler functions. For example, $ y(x) = x + 1 $ is the sum of the identity function $ x $ and the constant function $ 1. $ We will verify the



Sum Rule

\[ \frac{d}{dx}(y+z) = \frac{dy}{dx} + \frac{dz}{dx}. \]

Differentiating sums is pretty easy. Let's consider the average rate of change of the function $ y(x) + z(x) $ over the interval $ x_0 $ to $ x_0 + h. $

\begin{eqnarray*} \frac{(y(x_0 + h) + z(x_0 + h)) - (y(x_0) + z(x_0))}{h} & = & \frac{(y(x_0 + h) - y(x_0)) + (z(x_0 + h) - z(x_0))}{h} \\ & = & \frac{ y(x_0 + h) - y(x_0) }{h} + \frac{z(x_0 + h) - z(x_0)}{h}. \end{eqnarray*}

This shows that the average rate of change of the sum is just the sum of the average rates of change of $ y(x) $ and $ z(x). $

As $ h $ becomes very small, the two average rates of change are very close to the derivatives $ y^\prime(x_0) $ and $ z(x_0). $ This shows us that

\[ \frac{d(y+z)}{dx} = \frac{dy}{dx} + \frac{dz}{dx}. \]



Example:
Consider the function $ y(x) = x + 1. $ We know that the derivative

\[ \frac{dy}{dx} = \frac{d}{dx}(x) + \frac{d}{dx}(1) = 1 + 0 = 1; \]
that is, the derivative is equal to 1 at every point.

This is to be expected since the graph of this function is the straight line through $ (0,1) $ with slope one. Again, the tangent line at any point will be the same graph and so the derivative will be 1 at any point.