Continuity and Differentiability Miscellaneous Exercise contains a variety of questions based on the concepts covered throughout the chapter. Along with continuity, differentiability, derivatives of composite functions, parametric differentiation and second order derivatives, this exercise also introduces some special applications of differentiation.
One such concept is the derivative of a function with respect to another function. If both u and v are functions of x, then the derivative of u with respect to v is given by:
\displaystyle \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}
The exercise also includes a question on the differentiation of determinants and an application of finding the second order derivative in parametric form. These topics extend the standard rules of differentiation and help in understanding their wider applications.
In this exercise, we will apply the concepts learned in the chapter to solve a mixed set of NCERT questions and strengthen our understanding of Continuity and Differentiability.
Key Concepts
1. Derivative of One Function with Respect to Another
If both u and v are functions of x, then the derivative of u with respect to v is given by:
\displaystyle \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}
Example: If u=\sin x and v=\cos x, then
\displaystyle \frac{du}{dv}=\frac{\cos x}{-\sin x}=-\cot x
2. Second Derivative in Parametric Form
If x and y are expressed in terms of a parameter t, then
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
The second derivative is obtained by differentiating \frac{dy}{dx} with respect to t and then dividing by \frac{dx}{dt}.
\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
3. Differentiation of Determinants
If the entries of a determinant are differentiable functions of x, then the derivative of the determinant is obtained by differentiating one row(column) at a time while keeping the remaining rows(columns) unchanged and adding the resulting determinants.
For a determinant of order 3,
\displaystyle \frac{d}{dx} \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = \begin{vmatrix} a_1'&a_2'&a_3'\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} + \begin{vmatrix} a_1&a_2&a_3\\ b_1'&b_2'&b_3'\\ c_1&c_2&c_3 \end{vmatrix} + \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1'&c_2'&c_3' \end{vmatrix}
This result is analogous to the product rule of differentiation and is useful when the entries of the determinant depend on the same variable.
Note: If a row (or column) contains only constants, its derivative is zero. Therefore, only those rows (or columns) that contain functions of x contribute while differentiating a determinant.
In many problems, it is often simpler to first evaluate the determinant and then differentiate the resulting expression using the standard rules of differentiation.
4. Revision of Differentiation Rules
The miscellaneous exercise combines questions from different topics of the chapter. The following rules are frequently used:
- Chain Rule
- Product Rule
- Quotient Rule
- Parametric Differentiation
- Second Order Derivatives
- Derivatives of Trigonometric, Exponential and Logarithmic Functions
Tip: The Miscellaneous Exercise contains questions based on a variety of concepts covered in this chapter. Before starting a solution, identify the appropriate method and apply the relevant differentiation rules carefully. Regular practice of such mixed questions will improve both conceptual understanding and problem-solving speed.
Example: Continuity and Differentiability Miscellaneous
Example. Differentiate \sin^2 x with respect to e^{\cos x}.
Solution
Let
u=\sin^2 x and v=e^{\cos x}
Using the formula
\displaystyle \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}
we first find \frac{du}{dx} and \frac{dv}{dx}.
Differentiating u=\sin^2 x w.r.t. x, we get
\displaystyle \frac{du}{dx}=2\sin x\cos x
Differentiating v=e^{\cos x} w.r.t. x, we get
\displaystyle \frac{dv}{dx}=e^{\cos x}(-\sin x)
\displaystyle =-e^{\cos x}\sin x
Therefore,
\displaystyle \frac{du}{dv}=\frac{2\sin x\cos x}{-e^{\cos x}\sin x}
\displaystyle =-\frac{2\cos x}{e^{\cos x}}
\displaystyle =-2e^{-\cos x}\cos x
Hence,
\boxed{\displaystyle \frac{d(\sin^2 x)}{d(e^{\cos x})}=-2e^{-\cos x}\cos x}
Let us now solve all the NCERT questions step by step in the miscellaneous exercise of Continuity and Differentiability.
Question 1: Continuity and Differentiability Miscellaneous
1. Differentiate (3x^2-9x+5)^9 w.r.t. x.
Solution
Given,
y=(3x^2-9x+5)^9
Let
u=3x^2-9x+5 \qquad ...(1)
Then
y=u^9 \qquad ...(2)
Differentiating (2) w.r.t. u, we get
\displaystyle \frac{dy}{du}=9u^8 \qquad ...(3)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{du}{dx}=6x-9 \qquad ...(4)
From (3) and (4), using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
\displaystyle =9u^8(6x-9)
Substituting u=3x^2-9x+5, we get
\displaystyle \frac{dy}{dx}=9(3x^2-9x+5)^8(6x-9)
\displaystyle =27(2x-3)(3x^2-9x+5)^8
Hence,
\boxed{\displaystyle \frac{dy}{dx}=27(2x-3)(3x^2-9x+5)^8}
Question 2: Continuity and Differentiability Misc.
2. Differentiate \sin^3 x+\cos^6 x w.r.t. x.
Solution
Given,
y=\sin^3 x+\cos^6 x
Differentiating both sides w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(\sin^3 x)+\frac{d}{dx}(\cos^6 x)
Using the Chain Rule,
\displaystyle \frac{d}{dx}(\sin^3 x)=3\sin^2 x\cos x
and
\displaystyle \frac{d}{dx}(\cos^6 x)=6\cos^5 x(-\sin x)
\displaystyle =-6\sin x\cos^5 x
Therefore,
\displaystyle \frac{dy}{dx}=3\sin^2 x\cos x-6\sin x\cos^5 x
\displaystyle =3\sin x\cos x(\sin x-2\cos^4 x)
Hence,
\boxed{\displaystyle \frac{dy}{dx}=3\sin x\cos x(\sin x-2\cos^4 x)}
You may already be following Maths Better for NCERT Solutions for the topics like
Likewise NCERT’s Continuity and Differentiability Miscellaneous Exercise for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s move on to the next question.
Question 3: Continuity and Differentiability
3. Differentiate (5x)^{3\cos 2x} w.r.t. x.
Solution
Given,
y=(5x)^{3\cos 2x} \qquad ...(1)
Taking logarithms on both sides of (1), we get
\log y=\log\left((5x)^{3\cos 2x}\right)
\displaystyle =3\cos 2x\,\log(5x) \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=\left(3\cos 2x\right)\frac{1}{x}+\log(5x)\frac{d}{dx}(3\cos 2x)
\displaystyle =\frac{3\cos 2x}{x}-6\sin 2x\,\log(5x)
Therefore,
\displaystyle \frac{dy}{dx}=y\left(\frac{3\cos 2x}{x}-6\sin 2x\,\log(5x)\right)
Substituting the value of y from (1), we get
\displaystyle \frac{dy}{dx}=(5x)^{3\cos 2x}\left(\frac{3\cos 2x}{x}-6\sin 2x\,\log(5x)\right)
Hence,
\boxed{\displaystyle \frac{dy}{dx}=(5x)^{3\cos 2x}\left(\frac{3\cos 2x}{x}-6\sin 2x\,\log(5x)\right)}
Question 4: Continuity and Differentiability Miscellaneous
4. Differentiate \sin^{-1}(x\sqrt{x}) w.r.t. x, where 0\le x\le 1.
Solution
Given,
y=\sin^{-1}(x\sqrt{x})
Let
u=x\sqrt{x}=x^{3/2} \qquad ...(1)
Then
y=\sin^{-1}u \qquad ...(2)
Differentiating (2) w.r.t. u, we get
\displaystyle \frac{dy}{du}=\frac{1}{\sqrt{1-u^2}} \qquad ...(3)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{du}{dx}=\frac{3}{2}x^{1/2}=\frac{3}{2}\sqrt{x} \qquad ...(4)
From (3) and (4), using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
\displaystyle =\frac{1}{\sqrt{1-u^2}}\cdot\frac{3}{2}\sqrt{x}
\displaystyle =\frac{3\sqrt{x}}{2\sqrt{1-x^3}}
Hence,
\displaystyle \frac{d}{dx}\left(\sin^{-1}(x\sqrt{x})\right)=\frac{3\sqrt{x}}{2\sqrt{1-x^3}}
This can also be written as (NCERT Answer):
\boxed{\displaystyle \frac{dy}{dx}=\frac{3}{2}\sqrt{\frac{x}{1-x^3}}}
Note: Since 0\le x\le 1, we have 1-x^3\ge 0. Therefore, \sqrt{1-x^3} is non-negative and no modulus sign is required while simplifying the derivative.
Question 5: Miscellaneous Exercise
5. Differentiate \displaystyle \frac{\cos^{-1}\left(\frac{x}{2}\right)}{\sqrt{2x+7}} w.r.t. x, where -2\lt x\lt 2.
Solution
Given,
\displaystyle y=\frac{\cos^{-1}\left(\frac{x}{2}\right)}{\sqrt{2x+7}}
Let
\displaystyle u=\cos^{-1}\left(\frac{x}{2}\right)\qquad ...(1)
and
\displaystyle v=(2x+7)^{-1/2}\qquad ...(2)
Then
\displaystyle y=uv \qquad ...(3)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{du}{dx}=-\frac{\frac{1}{2}}{\sqrt{1-\left(\frac{x}{2}\right)^2}}
\displaystyle =-\frac{1}{\sqrt{4-x^2}} \qquad ...(4)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{dv}{dx}=-\frac{1}{2}(2x+7)^{-3/2}\cdot 2
\displaystyle =-(2x+7)^{-3/2} \qquad ...(5)
From (3), using the Product Rule,
\displaystyle \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}
Using (4) and (5), we get
\displaystyle \frac{dy}{dx}=-\cos^{-1}\left(\frac{x}{2}\right)(2x+7)^{-3/2}-\frac{(2x+7)^{-1/2}}{\sqrt{4-x^2}}
Taking (2x+7)^{-3/2} common,
\displaystyle \frac{dy}{dx}=-(2x+7)^{-3/2}\left[\cos^{-1}\left(\frac{x}{2}\right)+\frac{2x+7}{\sqrt{4-x^2}}\right]
Hence,
\displaystyle \frac{dy}{dx}=-\frac{1}{(2x+7)^{3/2}}\left[\cos^{-1}\left(\frac{x}{2}\right)+\frac{2x+7}{\sqrt{4-x^2}}\right]
This can also be written as (NCERT Answer):
\boxed{\displaystyle -\left[\frac{1}{\sqrt{4-x^2}\,\sqrt{2x+7}}+\frac{\cos^{-1}\left(\frac{x}{2}\right)}{(2x+7)^{3/2}}\right]}
Note: Since -2\lt x\lt 2, we have 4-x^2\gt 0. Therefore, \sqrt{4-x^2} is positive and no modulus sign is required while simplifying \sqrt{1-\left(\frac{x}{2}\right)^2}.
Important: Questions in the Miscellaneous Exercise are based on different concepts from the chapter. Some require logarithmic differentiation, some involve parametric or implicit differentiation, while others use second derivatives or determinants. Therefore, carefully identify the technique required before starting the solution.
Question 6: Continuity and Differentiability Miscellaneous
6. Differentiate
\displaystyle \cot^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]
w.r.t. x, where 0\lt x\lt \frac{\pi}{2}.
Solution
Let
\displaystyle y=\cot^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right] \qquad ...(1)
Rationalizing the denominator inside the bracket, we get
\displaystyle \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}
\displaystyle =\frac{\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)^2}{(1+\sin x)-(1-\sin x)}
\displaystyle =\frac{2+2\sqrt{1-\sin^2 x}}{2\sin x}
\displaystyle =\frac{1+\cos x}{\sin x}
Since 0\lt x\lt \frac{\pi}{2}
\displaystyle \frac{1+\cos x}{\sin x}=\cot\frac{x}{2}
Substituting in (1), we get
\displaystyle y=\cot^{-1}\left(\cot\frac{x}{2}\right)
\displaystyle =\frac{x}{2}
Differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{1}{2}
Hence,
\boxed{\displaystyle \frac{d}{dx}\left[\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\right]=\frac{1}{2}}
Note: Since 0\lt x\lt \frac{\pi}{2}, we have \cos x\gt 0. Hence \sqrt{1-\sin^2x}=|\cos x|=\cos x, which allows the simplification \frac{1+\cos x}{\sin x}=\cot\frac{x}{2}.
Question 7: Continuity and Differentiability Miscellaneous
7. Differentiate (\log x)^{\log x} w.r.t. x, where x>1.
Solution
Given,
y=(\log x)^{\log x} \qquad ...(1)
Taking logarithms on both sides of (1), we get
\log y=\log x\cdot \log(\log x) \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\log(\log x)+\log x\cdot\frac{1}{\log x}\cdot\frac{1}{x}
\displaystyle =\frac{\log(\log x)}{x}+\frac{1}{x}
\displaystyle =\frac{1+\log(\log x)}{x}
Therefore,
\displaystyle \frac{dy}{dx}=y\cdot\frac{1+\log(\log x)}{x}
Substituting the value of y from (1), we get
\displaystyle \frac{dy}{dx}=\frac{(\log x)^{\log x}}{x}\left(1+\log(\log x)\right)
Hence,
\displaystyle \frac{dy}{dx}=\frac{(\log x)^{\log x}}{x}\left(1+\log(\log x)\right)
which is the same as the NCERT Answer:
\boxed{\displaystyle (\log x)^{\log x}\left[\frac{1}{x}+\frac{\log(\log x)}{x}\right]}
Note: Since x\gt 1, we have \log x\gt 0. Therefore, \log(\log x) is well-defined and logarithmic differentiation can be applied directly.
Question 8: Continuity and Differentiability Misc. Exercise
8. Differentiate \cos(a\cos x+b\sin x) w.r.t. x, where a and b are constants.
Solution
Given,
y=\cos(a\cos x+b\sin x)
Let
u=a\cos x+b\sin x \qquad ...(1)
Then
y=\cos u \qquad ...(2)
Differentiating (2) w.r.t. u, we get
\displaystyle \frac{dy}{du}=-\sin u \qquad ...(3)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{du}{dx}=-a\sin x+b\cos x \qquad ...(4)
From (3) and (4), using the Chain Rule,
\displaystyle \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}
\displaystyle =(-\sin u)(-a\sin x+b\cos x)
\displaystyle =\sin u(a\sin x-b\cos x)
Substituting u=a\cos x+b\sin x, we get
\displaystyle \frac{dy}{dx}=\sin(a\cos x+b\sin x)(a\sin x-b\cos x)
Hence,
\displaystyle \frac{dy}{dx}=\sin(a\cos x+b\sin x)(a\sin x-b\cos x)
This can also be written as (NCERT Answer):
\boxed{\displaystyle \frac{dy}{dx}=(a\sin x-b\cos x)\sin(a\cos x+b\sin x)}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you understand the concepts more visually.
Now, let’s proceed to the next question.
Question 9: Logarithmic Differentiation Miscellaneous
9. Differentiate (\sin x-\cos x)^{(\sin x-\cos x)} w.r.t. x, where \frac{\pi}{4}\lt x\lt \frac{3\pi}{4}.
Solution
Given,
y=(\sin x-\cos x)^{(\sin x-\cos x)} \qquad ...(1)
Taking logarithms on both sides of (1), we get
\log y=(\sin x-\cos x)\log(\sin x-\cos x) \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{1}{y}\frac{dy}{dx}=(\cos x+\sin x)\log(\sin x-\cos x)+(\sin x-\cos x)\frac{\cos x+\sin x}{\sin x-\cos x}
\displaystyle =(\cos x+\sin x)\log(\sin x-\cos x)+(\cos x+\sin x)
\displaystyle =(\cos x+\sin x)\Big[1+\log(\sin x-\cos x)\Big]
Therefore,
\displaystyle \frac{dy}{dx}=y(\cos x+\sin x)\Big[1+\log(\sin x-\cos x)\Big]
Substituting the value of y from (1), we get
\displaystyle \frac{dy}{dx}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)\Big[1+\log(\sin x-\cos x)\Big]
Hence,
\displaystyle \frac{dy}{dx}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)\Big[1+\log(\sin x-\cos x)\Big]
which is as per the NCERT Answer
or, this can also be written as:
\boxed{\displaystyle \frac{dy}{dx}=y(\cos x+\sin x)\Big[1+\log(\sin x-\cos x)\Big]}
Note: Since \frac{\pi}{4}\lt x\lt \frac{3\pi}{4}, we have \sin x-\cos x\gt 0. Therefore, \log(\sin x-\cos x) is well-defined.
Question 10: Continuity and Differentiability Miscellaneous
10. Differentiate x^x+x^a+a^x+a^a w.r.t. x, for some fixed a\gt 0 and x\gt 0
Solution
Given,
y=x^x+x^a+a^x+a^a
Differentiating both sides w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{d}{dx}(x^x)+\frac{d}{dx}(x^a)+\frac{d}{dx}(a^x)+\frac{d}{dx}(a^a)
Now,
\displaystyle \frac{d}{dx}(x^x)=x^x(1+\log x)
\displaystyle \frac{d}{dx}(x^a)=ax^{a-1}
\displaystyle \frac{d}{dx}(a^x)=a^x\log a
and
\displaystyle \frac{d}{dx}(a^a)=0
Therefore,
\displaystyle \frac{dy}{dx}=x^x(1+\log x)+ax^{a-1}+a^x\log a
Hence,
\boxed{\displaystyle \frac{dy}{dx}=x^x(1+\log x)+ax^{a-1}+a^x\log a}
Several questions in this exercise become much simpler after applying suitable algebraic or trigonometric identities. Before differentiating, always look for opportunities to simplify the given expression and choose the most appropriate method of differentiation.
Question 11: Continuity and Differentiability
11. Differentiate x^{x^2-3}+(x-3)^{x^2} w.r.t. x, where x\gt 3.
Solution
Let
u=x^{x^2-3} \qquad ...(1)
Taking logarithms on both sides, we get
\log u=(x^2-3)\log x \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{1}{u}\frac{du}{dx}=2x\log x+\frac{x^2-3}{x}
Therefore,
\displaystyle \frac{du}{dx}=x^{x^2-3}\left(2x\log x+\frac{x^2-3}{x}\right) \qquad ...(3)
Now let
v=(x-3)^{x^2} \qquad ...(4)
Taking logarithms on both sides, we get
\log v=x^2\log(x-3) \qquad ...(5)
Differentiating (5) w.r.t. x, we get
\displaystyle \frac{1}{v}\frac{dv}{dx}=2x\log(x-3)+\frac{x^2}{x-3}
Therefore,
\displaystyle \frac{dv}{dx}=(x-3)^{x^2}\left(2x\log(x-3)+\frac{x^2}{x-3}\right) \qquad ...(6)
Since
y=u+v
Differentiatinng w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} \qquad ...(7)
Using (3) and (6) in (7), we get
\displaystyle \frac{dy}{dx}=x^{x^2-3}\left(2x\log x+\frac{x^2-3}{x}\right)+(x-3)^{x^2}\left(2x\log(x-3)+\frac{x^2}{x-3}\right)
Hence, as given in the NCERT answers:
\displaystyle \frac{dy}{dx}=x^{x^2-3}\left(\frac{x^2-3}{x}+2x\log x\right)+(x-3)^{x^2}\left(\frac{x^2}{x-3}+2x\log(x-3)\right)
Note: Since x\gt 3, both x and x-3 are positive. Therefore, logarithmic differentiation can be applied directly to both terms.
Question 12: Continuity and Differentiability Misc. Exercise
12. Find \displaystyle \frac{dy}{dx} if y=12(1-\cos t),\; x=10(t-\sin t), where -\frac{\pi}{2}\lt t\lt \frac{\pi}{2}.
Solution
Given,
y=12(1-\cos t) \qquad ...(1)
x=10(t-\sin t) \qquad ...(2)
Differentiating (1) w.r.t. t, we get
\displaystyle \frac{dy}{dt}=12\sin t \qquad ...(3)
Differentiating (2) w.r.t. t, we get
\displaystyle \frac{dx}{dt}=10(1-\cos t) \qquad ...(4)
From (3) and (4),
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\displaystyle =\frac{12\sin t}{10(1-\cos t)}
\displaystyle \frac{dy}{dx}=\frac{6}{5}\cdot\frac{\sin t}{1-\cos t}
Using the half-angle formulae
\displaystyle \sin t=2\sin\frac{t}{2}\cos\frac{t}{2}
and
\displaystyle 1-\cos t=2\sin^2\frac{t}{2}
we get
\displaystyle \frac{dy}{dx}=\frac{6}{5}\cdot\frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}}
\displaystyle =\frac{6}{5}\cot\frac{t}{2}
Hence,
\boxed{\displaystyle \frac{dy}{dx}=\frac{6}{5}\cot\frac{t}{2}}
Question 13: Continuity and Differentiability Miscellaneous
13. Find \displaystyle \frac{dy}{dx} if y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}, where 0\lt x\lt 1.
Solution
Given,
y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}
Differentiating both sides w.r.t. x, we get
\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-\left(\sqrt{1-x^2}\right)^2}}\cdot\frac{d}{dx}\left(\sqrt{1-x^2}\right)
\displaystyle =\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{x^2}}\cdot\frac{1}{2}(1-x^2)^{-1/2}(-2x)
\displaystyle =\frac{1}{\sqrt{1-x^2}}-\frac{x}{|x|\sqrt{1-x^2}}
Since 0\lt x\lt 1, we have |x|=x. Therefore,
\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}
\displaystyle =0
Hence,
\boxed{\displaystyle \frac{dy}{dx}=0}
Note: Since 0\lt x\lt 1, both x and \sqrt{1-x^2} are positive. Hence \sqrt{x^2}=|x|=x, which simplifies the derivative to zero.
Alternative Method (Useful for MCQs): Let \theta=\sin^{-1}x. Then x=\sin\theta and
\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta
Since 0\lt x\lt 1, we have 0\lt\theta\lt\frac{\pi}{2}. Therefore,
\sin^{-1}\sqrt{1-x^2}=\sin^{-1}(\cos\theta)=\frac{\pi}{2}-\theta
Hence,
y=\theta+\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}
which is a constant. Therefore,
\displaystyle \frac{dy}{dx}=0.
Question 14: Continuity and Differentiability Miscellaneous
14. If x\sqrt{1+y}+y\sqrt{1+x}=0, where -1\lt x\lt 1, prove that
\displaystyle \frac{dy}{dx}=-\frac{1}{(1+x)^2}
Solution
Given,
x\sqrt{1+y}+y\sqrt{1+x}=0 \qquad ...(1)
Transposing the second term to the RHS, we get
x\sqrt{1+y}=-y\sqrt{1+x}
Squaring both sides,
we get, x^2(1+y)=y^2(1+x)
or x^2+x^2y=y^2+xy^2
or x^2-y^2+x^2y-xy^2=0
(x-y)(x+y)+xy(x-y)=0
i.e. (x-y)(x+y+xy)=0
Since -1\lt x\lt 1, we have \sqrt{1+x}\gt 0 and \sqrt{1+y}\gt 0. Hence, from (1), x and y must have opposite signs or both be zero. Therefore, x=y is not possible unless x=y=0. Thus,
x+y+xy=0
y(1+x)=-x
\displaystyle y=-\frac{x}{1+x} \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{dy}{dx}=-\frac{(1+x)-x}{(1+x)^2}
\displaystyle =-\frac{1}{(1+x)^2}
Hence proved that
\boxed{\displaystyle \frac{dy}{dx}=-\frac{1}{(1+x)^2}}
Question 15: Miscellaneous Continuity and Differentiability
15. If (x-a)^2+(y-b)^2=c^2, where c\gt 0, prove that
\displaystyle \frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}
is a constant independent of a and b.
Solution
Given,
(x-a)^2+(y-b)^2=c^2 \qquad ...(1)
Differentiating (1) w.r.t. x, we get
2(x-a)+2(y-b)\frac{dy}{dx}=0
(x-a)+(y-b)\frac{dy}{dx}=0
\displaystyle \frac{dy}{dx}=-\frac{x-a}{y-b} \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=\frac{-(y-b)+(x-a)\frac{dy}{dx}}{(y-b)^2}
Using (2),
\displaystyle \frac{d^2y}{dx^2}=\frac{-(y-b)-\frac{(x-a)^2}{y-b}}{(y-b)^2}
\displaystyle =-\frac{(y-b)^2+(x-a)^2}{(y-b)^3}
Using (1),
\displaystyle \frac{d^2y}{dx^2}=-\frac{c^2}{(y-b)^3} \qquad ...(3)
Also, from (2),
\displaystyle \left(\frac{dy}{dx}\right)^2=\frac{(x-a)^2}{(y-b)^2}
Therefore,
\displaystyle 1+\left(\frac{dy}{dx}\right)^2=\frac{(y-b)^2+(x-a)^2}{(y-b)^2}
\displaystyle =\frac{c^2}{(y-b)^2}
\displaystyle \left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}=\frac{c^3}{(y-b)^3} \qquad ...(4)
From (3) and (4),
\displaystyle \frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}=\frac{\frac{c^3}{(y-b)^3}}{-\frac{c^2}{(y-b)^3}}
\displaystyle =-c
Since -c is a constant and does not involve a or b, the given expression is a constant independent of a and b.
Hence proved.
Important: Many proof-based questions can be simplified by first obtaining a suitable expression for \frac{dy}{dx} and then using it to eliminate terms while finding \frac{d^2y}{dx^2}. Careful algebraic and trigonometric simplification often makes such proofs much easier.
Question 16: Continuity and Differentiability Miscellaneous
16. If \cos y=x\cos(a+y), where \cos a\ne \pm 1, prove that
\displaystyle \frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}
Solution
Given,
\cos y=x\cos(a+y) \qquad ...(1)
Differentiating (1) w.r.t. x, we get
-\sin y\frac{dy}{dx}=\cos(a+y)-x\sin(a+y)\frac{dy}{dx}
Collecting the terms containing \frac{dy}{dx},
\left[x\sin(a+y)-\sin y\right]\frac{dy}{dx}=\cos(a+y)
\displaystyle \frac{dy}{dx}=\frac{\cos(a+y)}{x\sin(a+y)-\sin y} \qquad ...(2)
From (1),
\displaystyle x=\frac{\cos y}{\cos(a+y)}
Substituting in (2), we get
\displaystyle \frac{dy}{dx}=\frac{\cos(a+y)}{\frac{\cos y\sin(a+y)}{\cos(a+y)}-\sin y}
\displaystyle =\frac{\cos^2(a+y)}{\cos y\sin(a+y)-\sin y\cos(a+y)}
Using the identity
\displaystyle \sin A\cos B-\cos A\sin B=\sin(A-B)
with A=y and B=a+y, we get
\displaystyle \sin y\cos(a+y)-\cos y\sin(a+y)=\sin(y-a-y)=\sin(-a)=-\sin a
Therefore,
\displaystyle \cos y\sin(a+y)-\sin y\cos(a+y)=\sin a
Substituting above,
\displaystyle \frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}
Hence proved that
\boxed{\displaystyle \frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}}
Question 17: Second Order Parametric Differentiation
17. If x=a(\cos t+t\sin t) and y=a(\sin t-t\cos t), find \displaystyle \frac{d^2y}{dx^2}.
Solution
Given,
x=a(\cos t+t\sin t) \qquad ...(1)
y=a(\sin t-t\cos t) \qquad ...(2)
Differentiating (1) w.r.t. t, we get
\displaystyle \frac{dx}{dt}=a(-\sin t+\sin t+t\cos t)=at\cos t \qquad ...(3)
Differentiating (2) w.r.t. t, we get
\displaystyle \frac{dy}{dt}=a(\cos t-\cos t+t\sin t)=at\sin t \qquad ...(4)
From (3) and (4),
\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\displaystyle =\frac{at\sin t}{at\cos t}
\displaystyle =\tan t \qquad ...(5)
Now,
\displaystyle \frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(\tan t)=\sec^2 t
Using the formula
\displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}
we get
\displaystyle \frac{d^2y}{dx^2}=\frac{\sec^2 t}{at\cos t}
\displaystyle =\frac{1}{at\cos^3 t}
Hence,
\displaystyle \frac{d^2y}{dx^2}=\frac{1}{at\cos^3 t}
This can also be written as (NCERT Answer):
\boxed{\displaystyle \frac{d^2y}{dx^2}=\frac{\sec^3 t}{at}}
Question 18: Continuity and Differentiability Miscellaneous
18. If f(x)=|x|^3, show that f''(x) exists for all real x and find it.
Solution
We know that,
|x|^3=\begin{cases}x^3,&x\ge 0\\-x^3,&x\lt 0\end{cases}
Now, for x\gt 0,
f'(x)=\frac{d}{dx}(x^3)=3x^2
And for x\lt 0,
f'(x)=\frac{d}{dx}(-x^3)=-3x^2
Also for x=0,
\displaystyle f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{|h|^3}{h}
\displaystyle =\lim_{h\to 0}|h|^2=0
Hence,
f'(x)=\begin{cases}3x^2,&x\ge 0\\-3x^2,&x\lt 0\end{cases}\qquad ...(1)
Differentiating (1), we get
f''(x)=\begin{cases}6x,&x\gt 0\\-6x,&x\lt 0\end{cases}
Now at x=0,
\displaystyle f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h}
\displaystyle =\lim_{h\to 0}\frac{f'(h)}{h}
For h\gt 0,
\displaystyle \lim_{h\to 0^+}\frac{f'(h)}{h}=\lim_{h\to 0^+}\frac{3h^2}{h}=\lim_{h\to 0^+}3h=0
For h\lt 0,
\displaystyle \lim_{h\to 0^-}\frac{f'(h)}{h}=\lim_{h\to 0^-}\frac{-3h^2}{h}=\lim_{h\to 0^-}(-3h)=0
Since LHD = RHD,
f''(0)=0
Therefore,
f''(x)=\begin{cases}6x,&x\gt 0\\0,&x=0\\-6x,&x\lt 0\end{cases}
Since
|x|=\begin{cases}x,&x\ge 0\\-x,&x\lt 0\end{cases}
we get
f''(x)=6|x|
Hence,
\boxed{f''(x)=6|x|}
Thus, f''(x) exists for all real x.
Question 19: Continuity and Differentiability Miscellaneous
19. Using the fact that \sin(A+B)=\sin A\cos B+\cos A\sin B and differentiation, obtain the sum formula for cosines.
Solution
We know that
\sin(A+B)=\sin A\cos B+\cos A\sin B \qquad ...(1)
Assume that A and B are functions of a variable x. Differentiating both sides of (1) w.r.t. x, we get
\displaystyle \cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)=\frac{d}{dx}(\sin A\cos B)+\frac{d}{dx}(\cos A\sin B)
Using the Product Rule,
\displaystyle \cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)=\cos A\cos B\frac{dA}{dx}-\sin A\sin B\frac{dB}{dx}-\sin A\sin B\frac{dA}{dx}+\cos A\cos B\frac{dB}{dx}
\displaystyle =(\cos A\cos B-\sin A\sin B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)
Therefore,
\displaystyle \cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)=(\cos A\cos B-\sin A\sin B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)
Cancelling the common factor \displaystyle \left(\frac{dA}{dx}+\frac{dB}{dx}\right), we get
\displaystyle \cos(A+B)=\cos A\cos B-\sin A\sin B
Hence, the sum formula for cosines is
\boxed{\displaystyle \cos(A+B)=\cos A\cos B-\sin A\sin B}
Question 20: Miscellaneous Exercise
20. Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.
Solution
Consider the function
f(x)=|x^2-1|
Since x^2-1 is continuous for all real x and the modulus function preserves continuity, f(x) is continuous for all real x.
Now,
f(x)=\begin{cases}x^2-1,&x\le -1\text{ or }x\ge 1\\1-x^2,&-1\lt x\lt 1\end{cases}
Therefore,
f'(x)=\begin{cases}2x,&x\lt -1\text{ or }x\gt 1\\-2x,&-1\lt x\lt 1\end{cases}
At x=1,
\displaystyle \text{LHD}=\lim_{x\to 1^-}f'(x)=\lim_{x\to 1^-}(-2x)=-2
\displaystyle \text{RHD}=\lim_{x\to 1^+}f'(x)=\lim_{x\to 1^+}(2x)=2
Since LHD ≠ RHD, f(x) is not differentiable at x=1.
Similarly, at x=-1,
\displaystyle \text{LHD}=\lim_{x\to -1^-}f'(x)=\lim_{x\to -1^-}(2x)=-2
\displaystyle \text{RHD}=\lim_{x\to -1^+}f'(x)=\lim_{x\to -1^+}(-2x)=2
Since LHD ≠ RHD, f(x) is not differentiable at x=-1.
For all other values of x, f(x) is differentiable.
Thus, f(x)=|x^2-1| is continuous everywhere but not differentiable at exactly two points, namely x=-1 and x=1.
Hence, such a function does exist.
Important: While finding second order derivatives, all the standard rules of differentiation such as the Product Rule, Quotient Rule and Chain Rule continue to apply. Therefore, a strong understanding of first derivatives is essential.
Question 21: Differentiation of Determinants
21. If
\displaystyle y=\begin{vmatrix}f(x)&g(x)&h(x)\\ l&m&n\\ a&b&c\end{vmatrix}
prove that
\displaystyle \frac{dy}{dx}=\begin{vmatrix}f'(x)&g'(x)&h'(x)\\ l&m&n\\ a&b&c\end{vmatrix}
Solution
Given,
\displaystyle y=\begin{vmatrix}f(x)&g(x)&h(x)\\ l&m&n\\ a&b&c\end{vmatrix}
Using the rule for differentiation of determinants, we differentiate one row at a time and add the resulting determinants.
Since the second and third rows contain only constants, their derivatives are zero.
Therefore, only the first row contributes.
\displaystyle \frac{dy}{dx}=\begin{vmatrix}f'(x)&g'(x)&h'(x)\\ l&m&n\\ a&b&c\end{vmatrix}
Hence proved.
Alternative Method: Expand the determinant along the first row and then differentiate term-by-term. This also gives the same result.
Expanding the determinant along the first row, we get
\displaystyle y=f(x)\begin{vmatrix}m&n\\ b&c\end{vmatrix}-g(x)\begin{vmatrix}l&n\\ a&c\end{vmatrix}+h(x)\begin{vmatrix}l&m\\ a&b\end{vmatrix}
Since l,m,n,a,b and c are constants, differentiating w.r.t. x, we get
\displaystyle \frac{dy}{dx}=f'(x)\begin{vmatrix}m&n\\ b&c\end{vmatrix}-g'(x)\begin{vmatrix}l&n\\ a&c\end{vmatrix}+h'(x)\begin{vmatrix}l&m\\ a&b\end{vmatrix}
This is precisely the expansion of the determinant
\displaystyle \begin{vmatrix}f'(x)&g'(x)&h'(x)\\ l&m&n\\ a&b&c\end{vmatrix}
along its first row.
Therefore,
\boxed{\displaystyle \frac{dy}{dx}=\begin{vmatrix}f'(x)&g'(x)&h'(x)\\ l&m&n\\ a&b&c\end{vmatrix}}
Hence proved.
Note: While differentiating a determinant, differentiate one row (or column) at a time and add the resulting determinants. If a row (or column) contains only constants, its derivative is zero and the corresponding determinant vanishes. In the above question, only the first row contains functions of x, so only that row contributes to \frac{dy}{dx}.
Question 22: Continuity and Differentiability Miscellaneous
22. If y=e^{a\cos^{-1}x}, where -1\le x\le 1, show that
\displaystyle (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-a^2y=0
Solution
Given,
y=e^{a\cos^{-1}x} \qquad ...(1)
Differentiating (1) w.r.t. x, we get
\displaystyle \frac{dy}{dx}=e^{a\cos^{-1}x}\cdot a\left(-\frac{1}{\sqrt{1-x^2}}\right)
\displaystyle =-\frac{ay}{\sqrt{1-x^2}} \qquad ...(2)
Differentiating (2) w.r.t. x, we get
\displaystyle \frac{d^2y}{dx^2}=-a\frac{d}{dx}\left[y(1-x^2)^{-1/2}\right]
\displaystyle =-a\left[(1-x^2)^{-1/2}\frac{dy}{dx}+y\cdot\frac{x}{(1-x^2)^{3/2}}\right]
Using (2),
\displaystyle \frac{d^2y}{dx^2}=-a\left[-\frac{ay}{1-x^2}+\frac{xy}{(1-x^2)^{3/2}}\right]
\displaystyle =\frac{a^2y}{1-x^2}-\frac{axy}{(1-x^2)^{3/2}} \qquad ...(3)
Multiplying (3) by (1-x^2), we get
\displaystyle (1-x^2)\frac{d^2y}{dx^2}=a^2y-\frac{axy}{\sqrt{1-x^2}}
From (2),
\displaystyle \frac{ay}{\sqrt{1-x^2}}=-\frac{dy}{dx}
Therefore,
\displaystyle (1-x^2)\frac{d^2y}{dx^2}=a^2y+x\frac{dy}{dx}
Rearranging, we get
\displaystyle (1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-a^2y=0
Hence proved.
Common Mistakes to Avoid
- Forgetting logarithmic differentiation: Expressions such as x^x, (\log x)^{\log x}, (\sin x-\cos x)^{(\sin x-\cos x)} and x^{x^2-3} should be handled using logarithmic differentiation. Taking logarithms first often simplifies the differentiation considerably.
- Ignoring the given domain: Conditions such as 0\lt x\lt 1, -2\lt x\lt 2 or 0\lt x\lt \frac{\pi}{2} are often provided to simplify expressions involving square roots, modulus signs and inverse trigonometric functions.
- Forgetting the Chain Rule: Functions such as \sin^{-1}(x\sqrt{x}), e^{a\cos^{-1}x}, \cos(a\cos x+b\sin x) and (5x)^{3\cos 2x} require repeated use of the Chain Rule.
- Making sign errors in inverse trigonometric functions: Special care is required while differentiating \cos^{-1}x and \cot^{-1}x, since negative signs are frequently missed.
- Not using suitable trigonometric identities: Many expressions simplify significantly using identities such as \sin t=2\sin\frac{t}{2}\cos\frac{t}{2} and 1-\cos t=2\sin^2\frac{t}{2}. Simplifying before differentiating often reduces the amount of work.
- Making mistakes in parametric differentiation: When x and y are given in terms of a parameter t, first find \frac{dy}{dt} and \frac{dx}{dt}, then use \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.
- Using the wrong formula for the second derivative of a parametric curve: Remember that \displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}, not \frac{d}{dt}\left(\frac{dy}{dx}\right) alone.
- Forgetting differentiation with respect to another function: In questions asking to differentiate one function with respect to another, use \displaystyle \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}} instead of differentiating directly.
- Making errors while differentiating determinants: Differentiate one row (or column) at a time and add the resulting determinants. Any row or column containing only constants contributes zero to the derivative.
- Not simplifying before differentiating: In proof-based and inverse trigonometric questions, suitable algebraic or trigonometric simplifications often make the differentiation much easier and reduce the chances of error.
Continue Learning
After completing the Miscellaneous Exercise of Continuity and Differentiability, you should now be familiar with continuity, differentiability, logarithmic differentiation, parametric differentiation, differentiation with respect to another function, second order derivatives and differentiation of determinants. You have also seen how the Product Rule, Quotient Rule, Chain Rule and implicit differentiation can be applied in a variety of problems involving algebraic, trigonometric, inverse trigonometric, exponential and logarithmic functions.
To strengthen your understanding further, make sure that you revise:
- Continuity and differentiability of functions
- Derivatives of algebraic, trigonometric and inverse trigonometric functions
- Derivatives of exponential and logarithmic functions
- Logarithmic differentiation
- The Chain Rule for composite functions
- The Product Rule and Quotient Rule
- Implicit differentiation
- Differentiation with respect to another function using \displaystyle \frac{dy}{du}=\frac{\frac{dy}{dx}}{\frac{du}{dx}}
- Parametric differentiation and the formula \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
- Second derivatives in parametric form
- Differentiation of determinants
- Simplification using algebraic and trigonometric identities
- Proof-based questions involving first and second derivatives
Explore More
Try solving a variety of problems involving continuity, differentiability, logarithmic differentiation, parametric differentiation, second order derivatives and differentiation of determinants on your own. Practise questions based on algebraic, trigonometric, inverse trigonometric, exponential and logarithmic functions. Regular practice will help you apply the Chain Rule, Product Rule, Quotient Rule and implicit differentiation correctly, simplify expressions efficiently and improve your overall problem-solving skills in differential calculus.
All the best and keep learning 👍
Discover more from Maths Better
Subscribe to get the latest posts sent to your email.
