Differentiation of Functions, standard derivatives with Examples of differentiation

Differentiation is a method to find rates of change. It is an important topic for the JEE exam. Derivative of a function y = f(x) of a variable x is the rate of change of y, with respect to the rate of change of x. This article helps you to learn the derivative of a function, standard derivatives, theorems of derivatives, differentiation of implicit functions and higher order derivatives, along with solved examples.

How to Differentiate a Function

The differentiation of a function is a way to show the rate of change of a function at a given point. For real-valued functions, it is the slope of the tangent line at a point on a graph.

The derivative of y with respect to x is defined as the change in y over the change in x, as the distance between x₀ and x₁ becomes infinitely small (infinitesimal). The derivative is often written as dy/dx.

In mathematical terms, if f is a real-valued function, and a is a point in its domain of definition, the derivative of f at a is defined by

\begin{array}{l} f^{'} (a) = lim_{h \to 0} \frac{f (a + h) - f (a)}{h} \end{array}

Standard Derivatives

\begin{array}{l} 1] d x (u^{n}) = n u^{n - 1} \frac{d u}{d x} \end{array}

\begin{array}{l} 2] \frac{d}{d x} (c) = 0, where c is a constant \end{array}

\begin{array}{l} 3] If y = F (u) where u = f (x) then \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x} \end{array}

(Chain rule or function of a function rule)

4) Derivatives of trigonometric functions

\begin{array}{l} \frac{d}{d x} (\sin u) = \cos u \frac{d u}{d x} \\ \frac{d}{d x} (\cos u) = - \sin u \frac{d u}{d x} \\ \frac{d}{d x} (\tan u) = \sec^{2} u \frac{d u}{d x} \\ \frac{d}{d x} (\sec u) = \sec u \tan u \frac{d u}{d x} \\ \frac{d}{d x} (c o s e c u) = - c o s e c u c o t u \frac{d u}{d x} \\ \frac{d}{d x} (c o t u) = - c o s e c^{2} u \frac{d u}{d x} \end{array}

5) Derivatives of inverse trigonometric functions

\begin{array}{l} \frac{d}{d x} (\sin^{- 1} u) = \frac{1}{\sqrt{1 - u^{2}}} \frac{d u}{d x}, - 1 < u < 1 \\ \frac{d}{d x} (\cos^{- 1} u) = \frac{- 1}{\sqrt{1 - u^{2}}} \frac{d u}{d x}, - 1 < u < 1 \\ \frac{d}{d x} (\tan^{- 1} u) = \frac{1}{1 + u^{2}} \frac{d u}{d x} \\ \frac{d}{d x} c o s e c^{- 1} u = - \frac{1}{| u | \sqrt{u^{2} - 1}} \frac{d u}{d x} | u | > 1 \\ \frac{d}{d x} (\sec^{- 1} u) = \frac{1}{| u | \sqrt{u^{2} - 1}} \frac{d u}{d x} | u | > 1 \\ \frac{d}{d x} (\cot^{- 1} u) = - \frac{1}{1 + u^{2}} \frac{d u}{d x} \end{array}

6) Exponential and logarithmic functions

\begin{array}{l} \frac{d}{d x} (e^{u}) = e^{u} \frac{d u}{d x} \\ \frac{d}{d x} (a^{u}) = a^{u} \ln a \frac{d u}{d x}, where a > 0, a \neq 1 \\ \frac{d}{d x} (\ln u) = \frac{1}{u} \frac{d u}{d x} \\ \frac{d}{d x} (\ln_{a} u) = \frac{1}{u \ln a} \frac{d u}{d x}, where a > 0, a \neq 1 \end{array}

7) Hyperbolic functions

\begin{array}{l} \frac{d}{d x} (\sinh u) = \cosh u \frac{d u}{d x} \\ \frac{d}{d x} (\cosh u) = \sinh u \frac{d u}{d x} \\ \frac{d}{d x} (\tanh u) = s e c h^{2} u \frac{d u}{d x} \\ \frac{d}{d x} (s e c h u) = - s e c h u t a n h u \frac{d u}{d x} \\ \frac{d}{d x} (c o s e c h u) = - c o s e c h u c o t h u \frac{d u}{d x} \\ \frac{d}{d x} (c o t h u) = - c o s e c h^{2} u \frac{d u}{d x} \end{array}

8) Inverse hyperbolic functions

\begin{array}{l} \frac{d}{d x} s i n h^{- 1} u = \frac{1}{\sqrt{1 + u^{2}}} \frac{d u}{d x} \\ \frac{d}{d x} c o s h^{- 1} u = \frac{1}{\sqrt{u^{2} - 1}} \frac{d u}{d x}, u > 1 \\ \frac{d}{d x} t a n h^{- 1} u = \frac{1}{1 - u^{2}} \frac{d u}{d x}, | u | < 1 \\ \frac{d}{d x} c o s e c h^{- 1} u = - \frac{1}{| u | \sqrt{u^{2} + 1}} \frac{d u}{d x} u \neq 0 \\ \frac{d}{d x} s e c h^{- 1} u = - \frac{1}{u \sqrt{1 - u^{2}}} \frac{d u}{d x} 0 < u < 1 \\ \frac{d}{d x} c o t h^{- 1} u = \frac{1}{1 - u^{2}} \frac{d u}{d x}, | u | > 1 \end{array}

Some Standard Substitution

Expression Substitution

Simple tricks to solve complicated differential equations are listed below.

(i) If a function contains

\begin{array}{l} \sqrt{a^{2} - x^{2}} \end{array}

, then substitute x = a sinθ or x = a cosθ

(ii) If a function contains

\begin{array}{l} \sqrt{a^{2} + x^{2}} \end{array}

, then substitute x = a cotθ or x = a tan θ

(iii) If a function contains

\begin{array}{l} \sqrt{x^{2} + a^{2}} \end{array}

, then substitute x = a cosecθ or x = a sec θ

Theorems of Derivatives

Find some of the important theorem results below.

\begin{array}{l} \frac{d}{d x} [u \pm v] = \frac{d u}{d x} \pm \frac{d v}{d x} \\ \frac{d}{d x} u v = u \frac{d v}{d x} + v \frac{d u}{d x} (Product Rule) \\ \frac{d}{d x} \frac{u}{v} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}} (Quotient Rule) \\ \frac{d}{d x} (k f (x)) = k \frac{d}{d x} (f (x)), w h e r e k i s c o n s t a n t \end{array}

Example 1: Find dy/dx for y = x sinx log x

Solution:

\begin{array}{l} y^{‘} = 1 \times \sin x \times \log x + x \cos x \log x + x \sin x \times \frac{1}{x} = \sin x \log x + x \cos x \log x + \sin x \end{array}

Example 2: Find dy/dx for y = sin(x² + 1).

Solution:

\begin{array}{l} y^{‘} = \cos (x^{2} + 1) \times 2 x \end{array}

= 2x cos (x² + 1)

Differentiation of Implicit Function

Implicit differentiation, also known as the chain rule, differentiate both sides of an equation with two given variables by considering one of the variables as a function of the second variable. In short, differentiate the given function with respect to x and solve for dy/dx. Let us have a look at some examples.

Example 1: If x² + 2xy + y³ = 4, find dy/dx.

Solution: Differentiating both sides w.r.t. x, we get

\begin{array}{l} \frac{d}{d x} (x^{2}) + 2 \frac{d}{d x} (x y) + \frac{d}{d x} (y^{3}) = \frac{d}{d x} (4) \end{array}

\begin{array}{l} = 2 x + 2 x \frac{d y}{d x} + 2 y + 3 y^{2} . \frac{d y}{d x} = 0 \end{array}

\begin{array}{l} \Rightarrow \frac{d y}{d x} = \frac{- 2 (x + y)}{(2 x + 3 y^{2})} \end{array}

Example 2: Differentiate log sin x w.r.t

\begin{array}{l} \sqrt{\cos x} \end{array}

Solution: Let u = log sin x and v =

\begin{array}{l} \sqrt{\cos x} \end{array}

\begin{array}{l} \frac{d u}{d x} = \cot x & \frac{d v}{d x} = \frac{- \sin x}{2 \sqrt{\cos x}} \end{array}

\begin{array}{l} \frac{d u}{d x} = \frac{d u / d x}{d v / d x} = \frac{\cot x}{\frac{- \sin x}{2 \sqrt{\cos x}}} = - 2 \sqrt{\cos x} \cot x \cos e c (x) \end{array}

Higher Order Derivatives

The differentiation process can be continued up to the nth derivation of a function. Usually, we deal with the first-order and second-order derivatives of the functions.

dy/dx is the first derivative of y w.r.t x.

d²y/dx² is the second derivative of y w.r.t x.

Similarly, finding the third, fourth, fifth and successive derivatives of any function, say g(x), which are known as higher-order derivatives of g(x). The
nth order derivative numerical notation is gⁿ(x) or dⁿy/dxⁿ

Example: If

\begin{array}{l} y = e^{\tan^{- 1} x}, t h e n p r o v e t h a t (1 + x^{2}) \frac{d^{2} y}{d x^{2}} = (1 - 2 x) \frac{d y}{d x} \end{array}

Solution:

\begin{array}{l} y = e^{\tan^{- 1} x} \end{array}

\begin{array}{l} \frac{d y}{d x} = e^{\tan^{- 1} x} \frac{1}{1 + x^{2}} \end{array}

\begin{array}{l} = \frac{e^{t a n^{- 1} x}}{1 + x^{2}} \dots (i) \end{array}

\begin{array}{l} \frac{d^{2} y}{d x^{2}} = \frac{(1 + x^{2}) e^{t a n^{- 1} x} \frac{1}{1 + x^{2}} - 2 x e^{t a n^{- 1} x}}{(1 + x^{2})^{2}} \end{array}

\begin{array}{l} = \frac{(1 - 2 x) e^{t a n^{- 1} x}}{(1 + x^{2})^{2}} \end{array}

\begin{array}{l} \frac{d^{2} y}{d x^{2}} (1 + x^{2}) = \frac{(1 - 2 x) e^{t a n^{- 1} x}}{(1 + x^{2})} \end{array}

= (1-2x)dy/dx (from eqn (i))

Hence proved.

Video Lessons

Methods of Differentiation – JEE Solved Questions

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Important Theorems of Differentiation for JEE

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Frequently Asked Questions

What do you mean by differentiation in mathematics?

Differentiation is the process of finding the derivative of a function.

Give the product rule of differentiation.

Product rule: (d/dx) (uv) = u (dv/dx) + v (du/dx).

Give the quotient rule of differentiation.

Quotient rule: (d/dx)(u/v) = (v (du/dx) – u (dv/dx))/v².

What is the derivative of cot x w.r.t. x?

The derivative of cot x w.r.t.x = -cosec²x.

Differentiation of Functions

Related Topics:

How to Differentiate a Function

Standard Derivatives

Some Standard Substitution

Expression Substitution

Theorems of Derivatives

Differentiation of Implicit Function

Higher Order Derivatives

Video Lessons

Methods of Differentiation – JEE Solved Questions

Important Theorems of Differentiation for JEE

Frequently Asked Questions

What do you mean by differentiation in mathematics?

Give the product rule of differentiation.

Give the quotient rule of differentiation.

What is the derivative of cot x w.r.t. x?

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