# Differentiation is Linear

Differentiation is a process of finding the derivative of a function. Whenever we deal with Precalculus or Calculus we have to deal with differentiation. If we call differentiation as the heart of the calculus then it won't be wrong. Let's start with the Physical Significance of differentiation.

Let s = f( t ) -----(1), be any function which represents the distance travelled by any object in time ' t ', imagine the object is moving in a straight line, then after the time dt (small interval ). Let the distance covered be ds (small distance), then the equation becomes:

s + ds = f(t + dt)-----(2)

On subtracting  (1) from (2) we get:

ds = f(t + dt ) - f (t)

Further, to calculate the average velocity we differentiate with respect to t,

So, average velocity is,,

=> ds/dt = [f(t + dt ) - f (t) ] / dt,

When the time dt -> 0 , the average velocity will called as the instant velocity.

Velocity at time t = f' (t),

These are the few rules of linear Differential Equations. Now, have a look on some common laws of linear differentiation, which are as follows:

Law of power for linear differentiation:  d (x^n)/ dx = n . x^(n-1).

Law of addition for linear differentiation: d( u + v )/ dx = du/dx + dv/dx.

Law of subtraction for linear differentiation: d( u - v )/ dx = du/dx - dv/dx.

Law of product for linear differentiation: d( u * v )/ dx = (du/dx ) * ( dv/dx).

Even the Chain Rule works for the linear differential equations. Let's see how?

Let y = f(t) and t = g(x),

Then, dy/dx= ( dy/dt ) * ( dt/dx ).

We observe that these all rules work for the linear Functions. Let us take an example to understand the concept. Let function f(x) = (2x + 3 )^5?

Here, 2x +5 is in the form of a linear equation,

If we assume, 2x + 5 = s,

Then, y = s^5 and s = 2x +5,

dy/ds = 5 s^4,

Also ds/dx = 2,

Thus, we get,  dy/dx = (dy / ds)  * ( ds /dx),

= 5 * 2 * s^4,

= 10 * (2x + 3)^4,

Thus, we conclude that a linear equation is also differentiable by the simple rules of differentiation.

Note:

Relative rate of change for any function can be calculated using the standard formula, which is f'(x)/f(x), where f'(x) is the single derivative of f(x). We can understand this with the help of an example f(x) = x where rate of change will be x'/x = 1/x. The rate of change is positive if the dependent variable increases with time and negative if decreases.

Note:

To differentiate sum/difference of terms, differentiate the individual terms, add and subtract them together we must follow the steps shown below-

1.  Write the terms of expression individually.

2. Differentiate each term separately.

3. Add the results of differentiation of all the terms; this will be our final result.

Let us take an example to differentiate sum/difference of terms, differentiate the individual terms, add and subtract them together -

Example- Our expression is f(x) = x3 + 3x2 + 8x + 23

1.Differentiate x3 which gives d(x3)/dx = 3x2
2.Differentiate 3x2 which gives d(3x2)/dx = 6x
3.Differentiate 8x which gives d(8x)/dx = 8
4.Differentiate x3 which gives d(23)/dx = 0
5.Add results of terms after differentiation which is f'(x)=3x2 + 6x + 8

Note:

We Use power rule to differentiate radicals and reciprocals by first converting them to power functions. In this we simply have to adjust the power of a variable and then apply the standard formula f'(x) = nxn-1 for a function xn, where n can be in radical form (√) or “-” or “+” number or even a fraction.