# 3.2: Linearity of the Derivative

- Page ID
- 464

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

An operation is linear if it behaves "nicely'' with respect to multiplication by a constant and addition. The name comes from the equation of a line through the origin, \(f(x)=mx\), and the following two properties of this equation. First, \(f(cx)=m(cx)=c(mx)=cf(x)\), so the constant \(c\) can be "moved outside'' or "moved through'' the function \(f\). Second, \(f(x+y)=m(x+y)=mx+my= f(x)+f(y)\), so the addition symbol likewise can be moved through the function.

The corresponding properties for the derivative are:

\[ \left( cf(x) \right)' = {d\over dx}cf(x) = c {d\over dx} f(x) = cf'(x),\]

and

\[ \left( f(x)+g(x) \right)' = {d\over dx}(f(x)+g(x)) = {d\over dx} f(x)+{d\over dx} g(x) =f'(x)+g'(x).\]

It is easy to see, or at least to believe, that these are true by thinking of the distance/speed interpretation of derivatives. If one object is at position \(f(t)\) at time \(t\), we know its speed is given by \(f'(t)\). Suppose another object is at position \(5f(t)\) at time \(t\), namely, that it is always 5 times as far along the route as the first object. Then it "must'' be going 5 times as fast at all times.

The second rule is somewhat more complicated, but here is one way to picture it. Suppose a flat bed railroad car is at position \(f(t)\) at time \(t\), so the car is traveling at a speed of \(f'(t)\) (to be specific, let's say that \(f(t)\) gives the position on the track of the rear end of the car). Suppose that an ant is crawling from the back of the car to the front so that its position *on the car* is \(g(t)\) and its speed *relative to the car* is \(g'(t)\). Then in reality, at time \(t\), the ant is at position \(f(t)+g(t)\) along the track, and its speed is "obviously'' \(f'(t)+g'(t)\).

We don't want to rely on some more-or-less obvious physical interpretation to determine what is true mathematically, so let's see how to verify these rules by computation. We'll do one and leave the other for the exercises.

\[\eqalign{ {d\over dx}(f(x)+g(x)) &= \lim_{\Delta x\to 0} {f(x+\Delta x)+g(x+\Delta x) - (f(x)+g(x))\over \Delta x} \cr& = \lim_{\Delta x\to 0} {f(x+\Delta x)+g(x+\Delta x) - f(x)-g(x)\over \Delta x} \cr& = \lim_{\Delta x\to 0} {f(x+\Delta x)-f(x) +g(x+\Delta x) -g(x)\over \Delta x} \cr& = \lim_{\Delta x\to 0} \left({f(x+\Delta x)-f(x)\over \Delta x} +{g(x+\Delta x) -g(x)\over \Delta x}\right) \cr& = \lim_{\Delta x\to 0} {f(x+\Delta x)-f(x)\over \Delta x} + \lim_{\Delta x\to 0} {g(x+\Delta x) -g(x)\over \Delta x} \cr& =f'(x)+g'(x)\cr }\]

This is sometimes called the **sum rule** for derivatives.

Example 3.2.1

Find the derivative of \( f(x)=x^5+5x^2\).

**Solution**

We have to invoke linearity twice here:

\[f'(x) = {d\over dx}(x^5+5x^2) = {d\over dx}x^5 + {d\over dx}(5x^2) = 5x^4+5{d\over dx}(x^2) = 5x^4+5\cdot 2x^1 = 5x^4+10x.\]

Because it is so easy with a little practice, we can usually combine all uses of linearity into a single step. The following example shows an acceptably detailed computation.

Example 3.2.2

Find the derivative of \( f(x)=3/x^4-2x^2+6x-7\).

**Solution**

\[ f'(x) = {d\over dx}\left( {3\over x^4}-2x^2+6x-7\right) = {d\over dx}(3x^{-4}-2x^2+6x-7) = -12x^{-5}-4x+6.\]