Subject: Calculus

# Linearity Of Differentiation

When a function is expressed as a sum of individual products of a constant and a function, then we can't just differentiate that function directly. We still have no rule that can be applied to that yet. But intuitively, since there is a constant factor and sum involved, maybe there is a combination of this two simplest rules that can may define the derivative of the function stated above. In differential calculus, the most fundamental and important property of the derivative is the linearity of differentiation. It is a combination of the sum rule in differentiation and the constant factor rule in differentiation. Yes, this rule can be used to define the derivative of the function stated above. To better understand this, consider the kind of function we mentioned above.

### Illustration

Let f_{1} and f_{2} be differentiable functions at x and further assume that a and b are constants. Then a linear combination of these functions can be written as a linear function f,

f(x)=a\cdot f_{1}(x)+b\cdot f_{2}(x).
Now, we want to find the derivative of the linear function f, then
D_{x}[f(x)]=\frac{d}{dx}\left (a\cdot f_{1}(x)+b\cdot f_{2}(x)\right ).
By virtue of the sum rule in differentiation, the above function can be written as
f'(x)=\frac{d}{dx}[a\cdot f_{1}(x)]+\frac{d}{dx}[b\cdot f_{2}(x)],
then constant factor rule in differentiation suggests that it can have the following form,
f'(x)=a\frac{d}{dx}f_{1}(x)+b\frac{d}{dx}f_{2}(x).
Thus, we can alternatively write
\frac{d}{dx}f(x)=a\frac{d}{dx}f_{1}(x)+b\frac{d}{dx}f_{2}(x).
Now, notice that applying differentiation to the linear function f yields a linear function f'(x) composed of derivatives. This strongly tells us that differentiation is linear in nature.

### Example #1

Suppose that f(x) = g(x) and h(x), where g(x)=x^3+2x+1 and h(x)=x^2. Verify the notion of linearity of differentiation by showing that f'(x)=g'(x) + h'(x).

Since g(x)=x^3+2x+1 and h(x)=x^2, then we can write f(x)=x^3+x^2+2x+1. Differentiating f(x), we have

f'(x) = 3x^2+2x+2
.

On the other hand, the derivative of g(x) is just g'(x)=3x^2+2 and for h(x), h'(x)=2x. Adding g'(x) and h'(x), we get

g'(x) + h'(x) =3x^2+2 + 2x
= f'(x)
.

Thus, the linearity of differentiation is verified.

### Example #2

Let the function f(x) = g(x) and h(x), where g(x)=\cos{x}-\sin{x} and h(x)=\cos^{2}{x}. Verify that f'(x)=g'(x) + h'(x).

Since g(x)=\cos{x}-\sin{x} and h(x)=\cos^{2}{x}, then we can write f(x)=\cos{x}-\sin{x}+\cos^{2}{x}. Differentiating f(x), we have

f'(x) = -\sin{x}-\cos{x}-2\cos{x}\sin{x}
.

Meanwhile, the derivative of g(x) is just g'(x)=-\sin{x}-\cos{x} and for h(x), h'(x)=-2\cos{x}\sin{x}. Adding g'(x) and h'(x), we get

g'(x) + h'(x) =-(+\sin{x}+\cos{x}+2\cos{x}\sin{x})
f'(x)=g'(x) + h'(x)
.

Thus, the linearity of differentiation is verified.

### Example #3

If the values of a and b in the illustration above are expressed as functions of the differential variable x, namely a(x) and b(x), is the notion of linearity of differentiation still holds?

Yes of course. That's because differentiation is linear in nature as shown above.

### Example #4

From the scenario presented in example #3, show by direct differentiation that by replacing a and b by a(x) and b(x) the linearity still holds.

From

f(x)=a\cdot f_{1}(x)+b\cdot f_{2}(x)

we will have

f(x)=a(x)\cdot f_{1}(x)+b(x)\cdot f_{2}(x)

Letting f_1(x)\cdot a(x) = f_{a}(x) and f_{b}(x)=f_2(x)\cdot b(x), it follows that

f(x)= f_{a}(x)+f_{b}(x)

Differentiating the function directly, we have

\frac{d}{dx}f(x)= \frac{d}{dx}(f_{a}(x)+f_{b}(x))

By virtue of the sum rule in differentiation, the above function can be written as

\frac{d}{dx}f(x)=\frac{d}{dx}f_{a}(x)+\frac{d}{dx}f_{b}(x)

which just proves linearity of differentiation.

### Example #5

Name at least two differentiation rules that are used in verifying that differentiation is indeed a linear operation.

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