The derivative of a weighted sum

Definition 1

Consider to given function $u$ and $v$ , and two constant numbers $a$ and $b$ . The new function

f=a\cdot u+b\cdot v

is called a weighted sum (or a linear combination) of $u$ and $v$ . The "weights" $a$ and $b$ are also called coefficients.

The derivative of weighted sums is as follows:

Theorem 1

Assume a function $f$ can be written as a weighted sum of two other functions $u$ and $v$ , that is, $f=a\cdot u +b\cdot v$ . Then we have

Equation 1

f(x)=a\cdot u(x)+ b\cdot v(x) \overset{\prime}{\rightarrow} f^\prime(x)=a \cdot u^\prime(x)+b\cdot v^\prime(x)

Derivative of a weighted sum of functions

This is called the sum rule. It simply states that for forming $f^\prime$ , we have to replace $u$ and $v$ with their derivatives.

Proof

To prove the theorem, we first recall that for small $s$ (we normally use $h$ , but this letter is already used for the function $h$ ), we have

u'(x)=\lim_{h\rightarrow 0} \frac{u(x+h)-u(x)}{h}\approx \frac{u(x+h)-u(x)}{h}

v'(x)=\lim_{h\rightarrow 0} \frac{v(x+h)-v(x)}{h} \approx \frac{v(x+h)-v(x)}{h}

and

f(x)=\lim_{h\rightarrow 0} \frac{u(x+h)-v(x)}{h} \approx \frac{f(x+h)-f(x)}{h}

If we replace $f(x)$ with $a\cdot u(x)+b\cdot v(x)$ in the expression above, we get

\begin{array}{ll} f'(x) \approx \frac{f(x+h)-f(x)}{h} & = \frac{a\cdot u(x+h)+b\cdot v(x+h)- (a\cdot u(x)+b\cdot v(x))}{h} \\ & = \frac{a\cdot u(x+h)+b\cdot v(x+h)- a\cdot u(x)-b\cdot v(x))}{h}\\ & = \frac{a\cdot u(x+h)-a \cdot u(x) + b\cdot v(x+h)-b\cdot v(x))}{s}\\ & = a\cdot \frac{u(x+s)-u(x)}{h} - b\cdot \frac{v(x+h)-v(x)}{h} \\ & \approx a\cdot u'(x)+ b\cdot v'(x)\end{array}

And this completes the proof!

Example 1

The derivative of

f(x)=3x^4+6x^2 = \underbrace{3}_{a}\cdot \underbrace{x^4}_{u(x)}+\underbrace{6}_{b}\cdot \underbrace{x^2}_{v(x)}

f'(x)= \underbrace{3}_{a}\cdot \underbrace{4 x^3}_{u'(x)}+\underbrace{6}\cdot \underbrace{2 x^1}_{v'(x)}=12 x^3+12 x

Theorem 2

In particular we have

Equation 2

\begin{array}{lll} u(x)+v(x) &\overset{\prime}{\rightarrow}& u^\prime(x)+v^\prime(x)\\ a\cdot u(x) &\overset{\prime}{\rightarrow}& a\cdot u^\prime(x) \\ a\cdot u(x)+ b &\overset{\prime}{\rightarrow}& a \cdot u^\prime(x)\\ a\cdot u(x)+b\cdot v(x)+ c\cdot w(x) &\overset{\prime}{\rightarrow}& a\cdot u'(x)+b\cdot v'(x)+ c\cdot w'(x) \end{array}

The derivative of special weighted sums

Proof

$f(x)=u(x)+v(x) \rightarrow f'(x)=u'(x)+v'(x)$ (set $a=b=1$ )
$f(x)=a\cdot u(x) \rightarrow f'(x)=a\cdot u'(x)$ (set $b=0$ )
$f(x)=a\cdot u(x)+b = a\cdot u(x)+b\cdot 1= a\cdot u(x)+b = a\cdot u(x)+b\cdot x^0 \rightarrow f'(x)=a\cdot u'(x)+b\cdot 0\cdot x^{-1}=a\cdot u'(x)$
$f(x)=a\cdot u(x)+g(x)$ , where $g(x)=b\cdot v(x)+ c\cdot w(x)$ \rightarrow $f'(x)=a\cdot u'(x)+g'(x)=a\cdot u'(x)+b\cdot v'(x)+ c\cdot w'(x)$

Warning

The same law ist not correct for the multiplication of two functions. That is, if $f=u\cdot v$ , then in general it is $f^\prime \neq u^\prime \cdot v^\prime$ . See the problem below.

E.g. $u(x)=x^3$ and $v(x)=x^7$ , thus we have

f(x)=u(x)\cdot v(x)= x^3\cdot x^7=x^{10} \rightarrow f'(x)=10 x^9

But this is not

u'(x)\cdot v'(x) = 3x^2\cdot 7x^6 = 21x^{8}

So how can we determine the derivative if two functions are multiplied? There is a specific rule for this case, which we will learn about later. However, for the product of power functions it is straight forward. See the example below.

Example 2

Determine the derivative of the function $f(x)=x^2\cdot \sqrt{x}$

Solution

$f(x)=x^2\cdot\sqrt{x}=x^2\cdot x^{1/2}=x^{2.5} \rightarrow f'(x)=2.5 x^{1.5}$

Again, to find $h'(x)$ we cannot simply multiply the derivative of $x^2$ with the derivative of $\sqrt{x}$ . A common mistake!