CBSE Class 12 Mathematics Notes Chapter 5 Continuity and Differentiability

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Jasmine Grover

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Continuity and differentiability provide a deep insight into the basics of continuity, differentiability, and the relation between them.

  • The continuity of a function for a graph y = f(x) is a continuous wave.
  • Differentiation of a function is used to determine the change of the function concerning the change in the domain of the function.
  • Both the Continuity and Differentiability functions are complementary to each other.
  • The first function needs to be proved for continuity at a point x = a, and then it is proved for its differentiability at the point x = a. 
  • Continuous for a function is given by x = a, if Limx→af(x)=f(a).
  • Differentiability of the function at the same point x = a, if Limx→af′(a)=(f(a+h)−f(a))/h.

According to the CBSE Syllabus 2023-24, the chapter on Continuity and Differentiability comes under Unit 3 of Calculus. NCERT Class 12 Mathematics Unit Calculus holds a weightage of around 35 marks and includes Continuity and Differentiability topics. 

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Class 12 Mathematics Chapter 5 Notes – Continuity and Differentiability

Chain Rule

  • Chain Rule is a formula used in differential calculus that is used to find the derivative of composite functions.
  • It is also known as the composite function rule.
  • The chain rule formula for function y = f(x), where f(x) is a composite function such that x = g(t), is given as:

dy/dx =dy/du .du/dx

  • First, we need to differentiate the outer function while maintaining the consistency of the inner function.
  • Another chain rule formula is represented as follows:

y’ = d/dx ( f(g(x) ) = f’ (g(x)) · g’ (x)

Chain Rule

Chain Rule

Derivative of Inverse Trigonometric Functions

Inverse Trigonometric Functions dy/dx
y = sin-1(x) 1/√(1 – x2)
y = cos-1(x) -1/√(1 – x2)
y = tan-1(x) 1/(1 + x2)
y = cot-1(x) -1/(1 + x2)
y = sec-1(x) 1/[|x|√(x2 – 1)]
y = cosec-1(x) -1/[|x|√(x2 – 1)]

Derivatives of Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions

Derivative of Implicit Function

  • Implicit function is a type of function that is expressed in terms of dependent variables and independent variables.

  • It consists of multiple variables, and one of the variables is the function of another variable.
  • The function cannot be written in the form of y = f(x).
  • It has more than one solution for the given function.
  • To determine the derivative of the implicit function, first differentiate f(x, y) = 0, with respect to one independent variable x.
  • Next, find the dy/dx derivative of the expression by algebraically moving the variables.

Derivative of Implicit Function

Derivative of Implicit Function

Exponential and Logarithmic Functions

Exponential Function

  • The exponential function is a mathematical function that can be expressed in terms of y = f(x) = bx, where “x” is a variable and “b” is a constant.
  • b is also the base of the function such that b > 1. 
  • The change factor b follows the inequality condition 0 < b < 1. 
  • The domain and range of an exponential function is a set of all real numbers.

Logarithmic Functions

  • A logarithmic function is defined as the inverse of the exponential function.
  • The function is used when we need to work with really huge data sets.
  • It can mathematically be represented as follows:

x → logb x = y if by = x

  • If the value of base b is equal to 10, then it is called a common logarithm.
  • If the value of base b is equal to e, then it is called the natural logarithm.
  • The range of a logarithmic function is a set of all real numbers.

Exponential and Logarithmic Functions

Exponential and Logarithmic Functions

Derivative of Exponential and Logarithmic Functions

  • The derivative of exponential function ex with respect to x is written as:

d/dx(ex) = ex

  • The derivative of logarithmic function log x with respect to x is written as:

d/dx(log x) = 1/x

Derivative of Exponential and Logarithmic Functions

Derivative of Exponential and Logarithmic Functions

Derivative of Functions Expressed in Parametric Form

  • Parametric Form exists when a function y(x) is expressed in terms of the third variable.
  • The parametric form representation expressing the relation between x and y is written in the Form of x = f(t) and y = g(t).
  • It is used to represent the coordinates of any geometrical object, like curves and surfaces.
  • These functions are represented in two parts, namely first-order derivative and second-order derivative.

How to solve derivatives expressed in parametric Form

  • First, express the variables in terms of parameters, say t.
  • Now differentiate the variable with respect to t, such as dx/dt and dy/dt.
  • Use the formula to determine the required values.

Derivative of Functions Expressed in Parametric Form

Derivative of Functions Expressed in Parametric Form

Second Order Derivative

  • A second-order derivative is a type of derivative that is obtained from a first-order derivative.
  • The first-order derivative is used to express the slope of a function. 
  • The second-order derivative is used to define how the slope changes over the independent variable in the graph.
  • The most common example is velocity, which expresses the rate of change of an object’s position.
  • It is used to determine the maximum or minimum and inflexion point values. 

Condition for maximum or minimum and inflexion point values are as follows:

  • If the value of function f”(x) < 0, then f(x) has a local maximum at x.
  • If the value of function f”(x) > 0, then the f(x) has a local minimum at x.
  • If the value of function f”(x) = 0, then it is not possible to determine anything about x, which is the possible inflexion point.

Second Order Derivatives

Second Order Derivatives

There are Some important List Of Top Mathematics Questions On Continuity And Differentiability Asked In CBSE CLASS XII


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CBSE CLASS XII Related Questions

  • 1.
    Find : \[ \int \frac{2x+1}{\sqrt{x^2+6x}}\,dx \]


      • 2.

        Smoking increases the risk of lung problems. A study revealed that 170 in 1000 males who smoke develop lung complications, while 120 out of 1000 females who smoke develop lung related problems. In a colony, 50 people were found to be smokers of which 30 are males. A person is selected at random from these 50 people and tested for lung related problems. Based on the given information answer the following questions: 

        (i) What is the probability that selected person is a female? 
        (ii) If a male person is selected, what is the probability that he will not be suffering from lung problems? 
        (iii)(a) A person selected at random is detected with lung complications. Find the probability that selected person is a female. 
        OR 
        (iii)(b) A person selected at random is not having lung problems. Find the probability that the person is a male. 
         


          • 3.
            A line passing through the points \(A(1,2,3)\) and \(B(6,8,11)\) intersects the line \[ \vec r = 4\hat i + \hat j + \lambda(6\hat i + 2\hat j + \hat k) \] Find the coordinates of the point of intersection. Hence write the equation of a line passing through the point of intersection and perpendicular to both the lines.


              • 4.
                Evaluate : \[ \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}}(\sin|x|+\cos|x|)\,dx \]


                  • 5.
                    Find the general solution of the differential equation \[ y\log y\,\frac{dx}{dy}+x=\frac{2}{y}. \]


                      • 6.
                        Find the sub–interval of \((0,\pi)\) in which the function \[ f(x)=\tan^{-1}(\sin x-\cos x) \] is increasing and decreasing.

                          CBSE CLASS XII Previous Year Papers

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