Differentiation Formula

Differentiation Definition 

In mathematics, differentiation along with integration forms the basis and the most integral part of calculus. Differentiation is just a method of finding derivatives. The definition of the derivative is the rate of change, which means a particular parameter varies with respect to another parameter. These parameters are called functions. So in other words, the definition of the derivative is actually the rate of change of a particular function. One of the basic concepts of calculus mathematics is derivatives. Suppose there is a parameter or function y which is dependent on x. A small change in y can be written as dy and a small change in x can be written as dx. So the differentiation formula can be written as dy/dx it shows that the difference in y is divided by the difference in x. Here, d is not a variable, and therefore cannot be canceled out. Let us take an example, let's take velocity as one variable and time another variable. We know that velocity is dependent on time whereas time is not dependent on velocity. Therefore, velocity ‘v’ is a dependent variable, and time ‘t’ is an independent variable. If a moving body changes its velocity with respect to time then we denote this change as a change in velocity ‘dv’ divided by change in time ‘t’. This is the definition of the derivative because It allows us to determine the rate of change of velocity with respect to time which is acceleration. It also helps us to find the rate of change of variable x with respect to y. If we draw a graph of y against x then we can say that the graph is the gradient of the curve. The rate of change by doing the calculation of the ratio of change of the function y with respect to the change of the independent variable x is written dy/dx pronounced “dee y by dee x”.


Basic Differentiation Formula:

\[\frac{d}{dx}\] c = 0        (Here, c is a constant)

\[\frac{d}{dx}\] x = 1

\[\frac{d}{dx}\] x\[^{n}\]= n x\[^{n-1}\]

\[\frac{d}{dx}\]u ± v = \[\frac{d}{dx}\]u ±  \[\frac{d}{dx}\]v  

\[\frac{d}{dx}\]cu = c\[\frac{d}{dx}\]u  (c is constant)

ddx (uv) = udvdx + vdudx

\[\frac{d}{dx}\]uv = u\[\frac{d}{dx}\]v + v\[\frac{d}{dx}\]u   (Product Rule)

\[\frac{d}{dx}\] (\[\frac{f(x)}{g(x)}\]) = \[\frac{\frac{d}{dx}f(x)g(x)-f(x)\frac{d}{dx}g(x)}{g^{2}(x)}\] (Quotient Rule)


Derivative Formula of Log and Exponential Function:

\[\frac{d}{dx}\](e\[^{x}\]) = (e\[^{x}\])

\[\frac{d}{dx}\](ln(x)) = \[\frac{1}{x}\]

\[\frac{d}{dx}\](a\[^{x}\]) = a\[^{x}\]log a  

\[\frac{d}{dx}\](x\[^{x}\]) = x\[^{x}\](1+ln x)

\[\frac{d(log_{a}x)}{dx}\] = \[\frac{1}{x}\] x \[\frac{1}{\text{ln a}}\]


Derivative Formula of Trigonometric Functions

\[\frac{d}{dx}\] (sinx) = cos x

\[\frac{d}{dx}\] (cosx) = - sinx

\[\frac{d}{dx}\] (tanx) = sec²x

\[\frac{d}{dx}\] (cotx) = - cosec²x

\[\frac{d}{dx}\] (secx) = sec x tan x

\[\frac{d}{dx}\] (cosec x) = - cosec x cot x


Derivative Formula of Inverse Trigonometric Functions

\[\frac{d}{dx}\](sin\[^{-1}\]x) = \[\frac{1}{\sqrt{1-x^{2}}}\]

\[\frac{d}{dx}\](cos\[^{-1}\]x) = \[\frac{-1}{\sqrt{1-x^{2}}}\]

\[\frac{d}{dx}\](tan\[^{-1}\]x) = \[\frac{1}{\sqrt{1+x^{2}}}\]

\[\frac{d}{dx}\](cot\[^{-1}\]x) = \[\frac{-1}{\sqrt{1+x^{2}}}\]

\[\frac{d}{dx}\](sec\[^{-1}\]x) = \[\frac{1}{x\sqrt{x^{2}-1}}\]

\[\frac{d}{dx}\](cosec\[^{-1}\]x) = -\[\frac{1}{x\sqrt{x^{2}-1}}\]


Differentiation Rule

\[\frac{d}{dx}\] (f(x)g(x)) = f’(x)g(x) + f(x) g’(x)   (Product Rule)

\[\frac{d}{dx}\] (\[\frac{f(x)}{g(x)}\]) = \[\frac{f{}'(x)g(x) - f(x)g{}'(x)}{g(x)}\]  (Quotient Rule)

\[\frac{d(f(g(x)))}{dx}\] = f’ (g(x))g’(x)                 (Chain Rule)

f’(x) = \lim_{h\rightarrow 0} \[\frac{f(x+h)-f(x)}{h}\]   (First Derivative Rule)


Solved Examples


Example 1) For function 5\[^{x}\] + 2x\[^{-3}\] , find the derivative.


Solution 1) Calculating the derivative by applying the formula, \[\frac{d}{dx}\](x\[^{n}\])nx\[^{n-1}\] 

First, n = 5

First-term, 5x\[^{4}\]

n = -3

- 3x\[^{-4}\]

Second term -6x\[^{-4}\]

Therefore, on differentiating equation we get 5x\[^{4}\] - 6x\[^{-4}\]    


Example 2) Find the first derivative of a function f given by 

f(x) = (\[\sqrt{x}\] + 2x)(4x\[^{2}\] - 1)  


Solution 2) The function can be considered as a product of function U = \[\sqrt{x}\] + 2x 

and V = 4x\[^{2}\] - 1

Hence, calculating the derivative by using  product rule  

f’(x) = U’V + UV’

Now, on differentiating equation we get,

= (\[\frac{1}{\sqrt[2]{x}}\] + 2)(4x\[^{4}\] - 1) + (\sqrt{x} + 2x)(8x)

f’(x) = \[\frac{(1+2.2\sqrt{x})(4x^{2}-1)+2\sqrt{x}(8x)(\sqrt{x}+2x)}{2\sqrt{x}}\] 

f’(x) = \[\frac{4x^{2}-1+16x^{5/2}-4\sqrt{4}+16x^{2}+32x^{5/2}}{2\sqrt{x}}\] 

f’(x) = \[\frac{48x^{5/2}+20x^{2}-4x^{1/2}-1}{2\sqrt{x}}\] 

FAQ (Frequently Asked Questions)

Question 1: When Should we Use the Chain Rule and the Product Rule in Differentiation?

Answer: Both of the two rules are very useful for differentiating functions. The chain rule is used when we differentiate a 'function of a function', like f(g(x)) in general while the product rule is used when we are calculating the derivative of the two functions that are being multiplied together, like f(x)g(x) in general.

For example, f(x) = sin(3x) is the basic example of what we call a 'composite' function or a 'function of a function'. In this example, the two functions are as follows: function one takes x and then multiplies it by 3 while function two takes the sine of the answer given by function one. While differentiating equation we use the chain rule for these types of functions.

On the contrary, if the function in question would have been, say, f(x) = xcos(x), then we know that it is time to use the product rule. This is because there are two separate functions multiplied together: 'x' takes x but does nothing; 'cos(x)' takes the cosine of x. Note that they're separate functions: one that doesn't rely on the answer to the other!

Question 2: What Does the Quotient Rule Mean?

Answer: The quotient rule comes to play its role when a function is divided by another function and you need to find the derivative of it. It is one of the last main rules that we use to calculate derivatives.  

We can begin with f(x) = x / x². What can be the derivative of f(x)? Is it just the top or maybe the bottom? will it be the derivative of the top times or the derivative of the bottom? What might it be? Well, we can always simplify this into 1 / x, and then f`(x) would then be -1 / x² because 1 / x is the same as x⁻¹. Then we use the power rule. But is there a way that we can find without simplifying it? No, we won't be able to simplify it in every case. Think about sin(x) / x, or (x² + ln(x)) / cos(x). What about those cases? Perhaps, you must know the quotient rule.