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.