Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Advanced Differentiation

  • Last Updated : 14 Jul, 2021

Derivatives are used to measure the rate of change of any quantity. This process is called differentiation. It can be considered as a building block of the theory of calculus. Geometrically speaking, the derivative of any function at a particular point gives the slope of the tangent at that point of the function. There are many methods to solve and calculate the derivatives of the functions. The usual way is to calculate through the chain rule and power rule. In some cases, the functions become too complex to calculate. In such cases, there are certain methods that must be followed to make the calculations easy. 

Derivatives

Derivatives are defined as the rate of change of some variable/quantity which is observed. Suppose the variable or the quantity is governed by the equation f(x). The derivative of this function is denoted by \frac{df(x)}{dx} or F(x). There are certain functions whose derivatives must be kept in mind. These functions help in simplifying the calculation for the derivatives. The table given below shows the derivatives of some standard functions. 

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

Function Derivative
sin(x)cos(x)
cos(x)-sin(x)
xnnxn-1
tan(x)sec2(x)
cot(x)cosec2(x)

Sometimes, functions are made up of products or division of two or more functions. In such cases, the product and quotient rule is used. Consider a function h(x), 



Product Rule: If h(x) = f(x).g(x) 

h'(x) = f'(x)g(x) + f(x)g'(x) 

Quotient Rule: If h(x) = f(x)/g(x) 

 h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}

Chain Rule

Often the functions are made up of a composition of two or more standard functions. In such cases, the usual techniques to find the derivatives are not enough. In that case, the chain rule comes to the rescue. Consider f(x) and g(x) as two differentiable functions, assuming they make a new function h(x) = f(g(x)), which is basically a composition of these functions. In this case, the chain rule can be applied as mentioned below, 

h'(x) = \frac{dh(x)}{dx} \\      = \frac{df(g(x))}{dx} \\      = \frac{df(g(x))}{dx}\frac{dg}{dx}

Advanced Differentiation

There are some functions for which calculating the derivative is not easy. It requires some algebraic manipulation before calculating their derivatives. Examples of such functions are implicit functions, parametric functions, higher-order derivatives, and logarithmic derivatives. All these functions are composed of standard functions, but they are very complex to be solved by the simple chain rule or power rule.



Derivatives of Implicit Functions

Implicit functions are the functions that are of the form f(x, y) = 0. The usual functions that are encountered are of the form y = f(x). These functions are explicit functions because the definition of the y is expressed clearly and expressed in terms of x. In implicit functions, the relationship between x and y is not completely clear. Let’s see this method as an example. 

Question: Find the derivative of the function x2 + y2 = 4. 

Answer: 

Notice that the given function is an implicit function, 

x2 + y2 = 4

⇒ x2 + y2 – 4 = 0 

This function is in the form of f(x, y) = 0. 

 \frac{d(f(x,y))}{dx} = 0 \\ = \frac{d(x^2 + y^2  - 1)}{dx} = 0 \\ = \frac{d}{dx}(x^2) + \frac{d}{dx}y^2 - \frac{d}{dx}(1) = 0 \\ = 2x + 2y\frac{dy}{dx} = 0 \\ = \frac{dy}{dx} = \frac{-x}{y}

Derivatives of Parametric Functions

In the functions studied above, the functions had only two variables and a relationship was defined between them. Often, in some functions, the relationship between two variables is described using a third variable. These functions are called parametric functions. For example, x = f(t) and y = g(t) in one such function. In this function, the variables x and y are defined with respect to the variable “t”. In this case, calculating the rate of change of x with respect to y is not as clear. 

Question: Calculate the derivative for the following function. 



x = t2 and y = t3 + 3. 

Answer: 

This is an example of a parametric function. In this case, the derivatives are calculated with respect to the third variable for both the functions. 

x = t2

Differentiating with respect to “t”,

\frac{dx}{dt} = \frac{d}{dt}(t^2) \\ = \frac{dx}{dt} = 2t

\frac{dy}{dt} = \frac{d}{dt}(t^3 +3) \\ = \frac{dy}{dt} = 3t^2

Now these derivatives can be combined. 

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ = \frac{dy}{dx} = \frac{3t^2}{2t} \\ = \frac{dy}{dx} = \frac{3}{2}t

Higher-Order Derivatives

Just like the derivatives are defined as the rate of change of a quantity. Similarly, the rate of change of rate of change can also be calculated. Such derivatives that measure the rate of change of a derivative are called higher-order derivatives. For example, the rate of change of position is given by velocity, and acceleration measures the rate of change of velocity. For the function f(x), 



\frac{df}{dx} denotes a first-order derivative. Similarly, \frac{d^2f}{dx^2}  denotes the second-order derivative and so on. 

Sample Problems

Question 1: Find the second order of the derivative of f(x) = x3

Answer: 

The first order derivative of this function is given by, 

f(x) = x3

⇒ f'(x) = 3x2

Differentiating the function again, 

f”(x) = 6x 

Question 2: Find the derivative for the given function, 

f(x) = sin(x)cos(x)

Answer

Given: f(x) = sin(x)cos(x) 

The given function is the product of two functions, thus the product rule should be applied. 

f(x) = h(x)g(x) 

⇒f'(x) = h'(x)g(x) + h(x)g'(x)

Here, h(x) = sin(x) and g(x) = cos(x)

f'(x) = h'(x)g(x) + h(x)g'(x)

⇒ f'(x) = \frac{d}{dx}(sin(x))cos(x) + sin(x)\frac{d}{dx}(cos(x)

⇒f'(x) = sin^2(x) - cos^2(x)

Question 3: Find the derivative for the given function, 



f(x) = xsin(x)

Answer

Given: f(x) = xcos(x) 

The given function is the product of two functions, thus the product rule should be applied. 

f(x) = h(x)g(x) 

⇒f'(x) = h'(x)g(x) + h(x)g'(x)

Here, h(x) = x and g(x) = sin(x)

f'(x) = h'(x)g(x) + h(x)g'(x)

⇒ f'(x) = \frac{d}{dx}(x)sin(x) + x\frac{d}{dx}(sin(x))

⇒f'(x) = sin(x) + xcos(x)

Question 4: Find the derivative of the function x2 + 2x + y = 1. 

Answer: 

Notice that the given function is an implicit function, 

x2 + 2x + y = 1. 

⇒ x2 + 2x + y – 1 = 0 

This function is in the form of f(x, y) = 0. 

 \frac{d(f(x,y))}{dx} = 0 \\ = \frac{d(x^2 + y + 2x  - 1)}{dx} = 0 \\ = \frac{d}{dx}(x^2) + \frac{d}{dx}y + \frac{d}{dx}(2x) - \frac{d}{dx}(1) = 0 \\ = 2x + \frac{dy}{dx}  + 2= 0 \\ = \frac{dy}{dx} = -(2 + 2x)

Question 5: Calculate the derivative for the following function. 

x = t + 5and y = 3t3 + 6. 

Answer: 



This is an example of a parametric function. In this case, the derivatives are calculated with respect to the third variable for both the functions. 

x = t + 5

Differentiating with respect to “t”,

\frac{dx}{dt} = \frac{d}{dt}(t + 5) \\ = \frac{dx}{dt} = 1

\frac{dy}{dt} = \frac{d}{dt}(3t^3 +3) \\ = \frac{dy}{dt} = 9t^2

Now these derivatives can be combined. 

\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \\ = \frac{dy}{dx} = \frac{9t^2}{1} \\ = \frac{dy}{dx} = 9t^2

Question 6: Find the third order of the derivative of f(x) = ex + sin(x). 

Answer: 

The first order derivative of this function is given by, 

f(x) = ex + sin(x)

⇒ f'(x) = ex + cos(x)

Differentiating the function again, 

f”(x) = ex – sin(x)

Differentiating the function again, 

f”'(x) = ex – sin(x)




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!